DIOPHANTINE EQUATIONS WITH BINOMIAL COEFFICIENTS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS

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1 DIOPHANTINE EQUATIONS WITH BINOMIAL COEFFICIENTS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA Abstract Ths work presents a study of perturbatons of symmetrc Booean functons In partcuar, t estabshes a connecton between exponenta sums of these perturbatons and Dophantne equatons of the form n n x 0, 0 where x j beongs to some fxed bounded subset Γ of Z The concepts of trvay baanced symmetrc Booean functon and sporadc baanced Booean functon are extended to ths type of perturbatons An observaton made by Canteaut and Vdeau [4] for symmetrc Booean functons of fxed degree s extended To be specfc, t s proved that, excudng the trva cases, baanced perturbatons of fxed degree do not exst when the number of varabes grows Some sporadc baanced perturbatons are presented Fnay, a beautfu but unexpected dentty between perturbatons of two very dfferent symmetrc Booean functons s aso ncuded n ths work 1 Introducton The theory of Booean functons s a beautfu area of combnatorcs wth vast appcatons to many areas of mathematcs as we as outsde the dscpne Exampes ncude eectrca engneerng, the theory of error-correctng codes and cryptography In the modern era, effcent mpementatons of Booean functons wth many varabes s a chaengng probem due to memory restrctons of current technoogy Because of ths, symmetrc Booean functons are good canddates for effcent mpementatons However, symmetry s a too speca property and may mpy that these mpementatons are vunerabe to attacks For ths reason, we study perturbatons of symmetrc Booean functons These perturbatons, whch are the focus of [7], are not onger symmetrc Nevertheess, the symmetry of the underyng functon can be expoted n order to make fast cacuatons, to obtan recurrences, and, as t was done n [7], to obtan nformaton about the asymptotc behavor In penty of appcatons, especay the ones reated to cryptography, t s mportant for Booean functons to be baanced A baanced Booean functon s one for whch the number of zeros and the number of ones are equa n ts truth tabe Baancedness of Booean functons can be studed from the pont of vew of exponenta sums, as t s done s ths artce Ths pont of vew s n fact a very actve area of research For some exampes, pease refer to [1, 2, 5 8, 15, 18 20, 22] The study of baancedness of symmetrc Booean functons s connected to the probem of bsectng bnoma coeffcents A souton δ 0, δ 1,, δ n to the equaton n n 11 x 0, x { 1, 1}, 0 s sad to gve a bsecton of the bnoma coeffcents n, 0 n The frst detaed study of ths connecton was made by Mtche [17] Other studes ncude Jefferes [14] and Sarkar and Matra [21] In ths work, baancedness of the perturbatons consdered s nked to equaton 11, where the x s now e n a bounded subset Γ of Z nstead of n { 1, 1} The concept of trvay baanced symmetrc Booean functon and the concept of sporadc baanced symmetrc Booean functon whch was ntroduced n [14], are extended to these perturbatons Date: December 28, Mathematcs Subject Cassfcaton 05E05, 11T23, 11D04 Key words and phrases Bnoma Dophantne equatons, perturbatons of symmetrc Booean functons, exponenta sums, recurrences 1

2 2 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA A conjecture smar to the one presented n [10] for eementary symmetrc Booean functons seems to be true for the smpest type of the perturbatons consdered n ths study Aso, smar to the case of symmetrc Booean functons, computatons suggest that most baanced perturbatons are trvay baanced Ths ed us to study trvay baanced perturbatons n more deta: we showed that once a perturbaton of fxed degree s trvay baanced at one pont, then t s trvay baanced at nfntey many ponts In [4], Canteaut and Vdeau observed that, excudng the trva cases, baanced symmetrc Booean functons of fxed degree do not exst when the number of varabes grows ths was recenty proved n [12] Ths resut s extended to our perturbatons, that s, t s proved that, excudng the trva cases, baanced perturbatons of fxed degree do not exst when the number of varabes grows Therefore, the search for sporadc baanced perturbatons s of nterest It s n ths search that an strkng dentty between the perturbatons of two dfferent symmetrc Booean functons s used In partcuar, ths dentty aow us to obtaned two sporadc baanced perturbatons for the prce of one Ths artce s dvded as foows The next secton ncudes some premnares that are needed for the work presented n ths manuscrpt In secton 3, a beautfu but unexpected dentty between the perturbatons of two very dfferent symmetrc Booean functons s proved Ths dentty s ater used n secton 6 when the search of sporadc perturbatons s consdered In secton 4, the study of the nk between perturbatons and equaton 11 over bounded sets of ntegers s consdered Some of the known resuts about baancedness of symmetrc Booean functons and bsectons of bnoma coeffcents are extended Secton 5 presents a study of baanced perturbatons as the number of varabes grows It s n ths secton where the observaton of Canteaut and Vdeau s extended Fnay, as mentoned before, some exampes of sporadc perturbatons are presented n secton 6 2 Premnares Let F 2 be the bnary fed, F2 n {x 1,, x n x F 2, 1,, n}, and F X F X 1,, X n be a poynoma n n varabes over F 2 The exponenta sum assocated to F over F 2 s 21 SF 1 F x1,,xn x 1,,x n F 2 A Booean functon F s caed baanced f SF 0, e the number of zeros and the number of ones are equa n the truth tabe of F Ths property s mportant for some appcatons n cryptography Any symmetrc Booean functon s a near combnaton of eementary symmetrc poynomas Let σ n,k be the eementary symmetrc poynoma n n varabes of degree k For exampe, 22 σ 4,3 X 1 X 2 X 3 + X 1 X 4 X 3 + X 2 X 4 X 3 + X 1 X 2 X 4 Every symmetrc Booean functon can be dentfed wth an expresson of the form 23 σ n,k1 + σ n,k2 + + σ n,ks, where 1 k 1 < k 2 < < k s are ntegers For the sake of smpcty, the notaton σ n,[k1,,k s] s used to denote 23 For exampe, 24 σ 3,[2,1] σ 3,2 + σ 3,1 X 1 X 2 + X 3 X 2 + X 1 X 3 + X 1 + X 2 + X 3 It s not hard to show that f 1 k 1 < k 2 < < k s are fxed ntegers, then n 25 Sσ n,[k1,k 2,,k s] 1 k 1 + k n ks 0 Remark 21 Observe that the rght hand sde of 25 makes sense for n 1, whe the eft hand sde exsts for n k s Throughout the rest of the artce, Sσ n,[k1,k 2,,k s] shoud be nterpreted as the expresson on the rght hand sde, so t makes sense to tak about exponenta sums of symmetrc Booean functons wth ess varabes than ther degrees In [6], Castro and Medna used 25 to study exponenta sums of symmetrc poynomas from the pont of vew of nteger sequences As part of ther study, they showed that the sequence {Sσ n,[k1,,k s]} n N

3 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 3 satsfes the homogeneous near recurrence 2 r x n 1 1 r x n, 1 where r og 2 k s + 1 ths resut aso foows from [3, Th 31, p 248] and used ths resut to compute the asymptotc behavor Sσ n,[k1,,k s] as n To be specfc, 1 27 m n 2 n Sσ n,[k 1,,k s] c 0 k 1,, k s where 28 c 0 k 1,, k s 1 2 r 1 2 r 1 k ks 0 They used ths concept to show that a conjecture of Cusck, L and Stǎncǎ s true asymptotcay ths resut was recenty re-estabshed n [12] See [6] for more detas In the case of the eementary symmetrc poynoma, the same authors were abe to mprove 26 and reduced the degree of the homogeneous near recurrence wth nteger coeffcents that ts exponenta sums satsfy They dd ths by fndng the mnma homogeneous near recurrence wth nteger coeffcents that {Sσ n,k } satsfes To be specfc, et ɛn be defned as { 0, f n s a power of 2, 29 ɛn 1, otherwse Then, the foowng resut hods see [6] Theorem 22 Let k be a natura number and p k X be the characterstc poynoma assocated to the mnma near recurrence wth nteger coeffcents that {Sσ n,k } n N satsfes Let k 2 k/2 + 1 Express k as ts 2-adc expanson 210 k a a as, where the ast exponent s gven by a s og 2 k Then, s 211 p k X X 2 ɛk Φ 2 a +1X 1 In partcuar, the degree of the mnma near recurrence that {Sσ n,k } n N satsfes s equa to 2 k/2 +ɛk In ths artce, perturbatons of symmetrc Booean functons and ther connecton to soutons of 11 over some bounded set of ntegers are consdered Reca that σ n,k s the eementary symmetrc poynoma of degree k n the varabes X 1,, X n Suppose that j < n and et F X be a bnary poynoma n the varabes X 1,, X j the frst j varabes n X 1,, X n We are nterested n exponenta sums of poynomas of the form 212 σ n,[k1,,k s] + F X, where 1 k 1 < < k s Observe that perturbatons of the form 212 are not necessary symmetrc In [7], Castro and Medna showed that exponenta sums of perturbatons of the form 212 are reated to exponenta sums of symmetrc Booean functons va the foowng equaton 213 Sσ n,[k1,,k s] + F X where 214 C m F 1 j m C m F S x F 2 wth w 2xm m σ n j,[k1,,k s ] 1 F x, and w 2 x represents the Hammng weght of x, e the number of entres of x that are one,

4 4 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA Remark 23 There are three thngs to observe about equaton 213 Frst, t s cear that the vaue of m that s nsde the exponenta sum can be taken mod 2, snce ony the party matters Second, f k < 0, then the term σ n j,k does not exst and so t s not present n the equaton Fnay, n the case that k 0, the eementary poynoma σ n j,0 shoud be nterpreted as 1 Equaton 213 now mpes that exponenta sums of these type of perturbatons aso satsfy recurrence 26 Ths s used n the next secton when an dentty between perturbatons of two dfferent symmetrc Booean poynomas s estabshed It aso serves as the nk to equaton 11 over a bounded set of ntegers 3 Some perturbatons denttes In ths secton we estabsh a very beautfu dentty between perturbatons of two dfferent symmetrc Booean poynomas We start our dscusson wth a partcuar exampe The dea for dong ths s to get an nsght of what s behnd ths dentty A proof for the genera case w be provded ater n ths secton once our ntuton s sodfed Consder the two poynomas σ n,4 and σ n,5 and ther correspondng exponenta sums: n 31 Sσ n,4 1 4 n n and Sσ n,5 1 5 n 0 These sums seem smar, but they have very dfferent behavors For exampe, consder the expressons and 1 5, whch are the coeffcents of the bnoma numbers n the sums 31 As ranges through the non-negatve ntegers, both expressons n 32 are perodc wth perod ength 8 In fact, ther perods are gven by Observe that these perods dffer n ony two postons, but ths dfference has a bg effect on the behavor of Sσ n,4 and Sσ n,5 For nstance, Sσ n,4 0, e σ n,4 s baanced, whenever n 7 mod 8 In contrast, the exponenta sum Sσ n,5 s never zero Aso, the exponenta sum Sσ n,4 assumes negatve vaues for some vaues of n, but Sσ n,5 s aways postve These cams are evdent from the frst few vaues of both sequences: n Sσ n, Sσ n, Thus, t s cear that the sequences {Sσ n,4 } and {Sσ n,5 } have very dfferent behavor However, both of them can be atered to make them, not just smar, but equa up to a shft! The trck, n fact, s very smpe: just add the near poynoma X 1 to both symmetrc poynomas, σ n,4 and σ n,5, to get 33 Sσ n,4 + X 1 Sσ n+1,5 + X 1 For exampe, n Sσ n,4 + X Sσ n,5 + X Ths trck not ony works for σ n,4 and σ n,5, but aso for σ n,2k and σ n,2k+1 wth k a postve nteger Expcty, 34 Sσ n,2k + X 1 Sσ n+1,2k+1 + X 1 We now begn our proof of dentty 34 The buk of the proof rees on the fact that sequences of the form {Sσ n,k } and {Sσ n,k +F X} satsfy near recurrences wth nteger coeffcents To be more specfc, the dea to estabsh dentty 34 s to show that both sequences satsfy the same recurrence Once ths s done, then to show that dentty 34 hods, t s enough to show that both sequences have the same nta condtons We start our argument wth the foowng resut

5 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 5 Lemma 31 Let k > 1 and m 1 be fxed ntegers Consder the sequence m 35 S m σ n,k j j Then, sequence 35 satsfes the same homogeneous near recurrence as {Sσ n,k } j0 Proof The proof s by nducton on m Start wth the dentty 36 Sσ n,k Sσ n 1,k + Sσ n 1,k + σ n 1,k 1 Ths mpes that 37 Sσ n,k + σ n,k 1 Sσ n+1,k Sσ n,k, and so t s cear that {Sσ n,k + σ n,k 1 } satsfes the same recurrence as {Sσ n,k } Thus, the cam hods for m 1 Suppose now that the statement s true for a vaues of m 1 ess than some m > 1 The dentty 38 mpes 39 S Sσ n,k m m 0 m 1 0 m S m S m σ n,k Sσ n+m,k σ n m,k σ n m,k + S m 1 0 m S m m σ n m,k, σ n,k By nducton, each term of the sum on the rght hand sde of 39 satsfes the same recurrence as {Sσ n,k }, therefore the cam s aso true for m Ths concudes the proof The next step s to show that the sequences {Sσ n,k } and {Sσ n,k + F X} satsfy the same near recurrence wth nteger coeffcents Theorem 32 Let k > 1 and j be fxed ntegers and et F X be a bnary poynoma n the varabes X 1,, X j Suppose that k 2 as a1 + 1 The sequence 310 {Sσ n,k + F X} n N satsfes the homogeneous near recurrence whose characterstc poynoma s s 311 fx X 2 ɛk Φ 2 a +1X 1 Moreover, f F X s baanced then 310 satsfes the homogeneous near recurrence wth characterstc poynoma s 312 Φ 2 a +1X 1, whch s of degree k 1 2 as a1 Proof Reca that 313 Sσ n,k + F X where 314 C m F 1 1 j m m C m F S σ n j,k, x F j 2 wth w2xm 1 F x,

6 6 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA and w 2 x s the Hammng weght of x Thus, the frst cam s a drect consequence of Lemma 31 For the second cam, the fact that fx s the characterstc poynoma mpes that 315 Sσ n,k + F X d 0 2 n + d λ λ n λ 2,fλ0 for some constants d 0 and d λ s In [7], Castro and Medna showed that 316 d 0 c 0 k SF 2 j, where c 0 k s defned by 28 The resut foows The above resuts are suffcent to show that for k 1 an nteger, 317 Sσ n,2k + X 1 Sσ n+1,2k+1 + X 1 for every postve nteger n Observe that snce F X X 1 s a baanced poynoma, then Theorem 32 mpes that the sequences {Sσ n,2k + X 1 } and {Sσ n+1,2k+1 + X 1 } satsfy the same near recurrence of order 2k Therefore, to show that both sequences are equa, t s suffcent to show that ther frst 2k vaues concde Let and Note that fn, k Sσ n,2k + X 1 Sσ n 1,2k Sσ n 1,2k + σ n 1,2k 1 n 1 1 2k j 1 1 2k 1 j n 1 j j0 gn, k Sσ n+1,2k+1 + X 1 Sσ n,2k+1 Sσ n,2k+1 + σ n,2k n 1 2k+1 j 1 1 2k j n j j0 f1, k 0 g1, k, f2, k 0 g2, k, f2k 1, k 0 g2k 1, k, f2k, k 2 g2k, k In other words, {fn, k} n1 and {gn, k} n1 satsfy the same recurrence of order 2k wth the same nta condtons Therefore, fn, k gn, k, Sσ n,2k + X 1 Sσ n+1,2k+1 + X 1 for every n and k Ths dscusson aso mpes the foowng dentty of bnoma sums Coroary 33 Let k and n be postve ntegers Then, n 1 2k k n 1 n 1 2k 1 1 2k 1 n Therefore, even though {Sσ n,2k } and {Sσ n,2k+1 } are dfferent sequences, they can be atered n such a way to produce an equaty up to a shft n the number of varabes, e 318 Sσ n,2k + X 1 Sσ n+1,2k+1 + X 1

7 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 7 Equaton 318 eads to the foowng queston: For whch Booean poynomas F X n j varabes j fxed does the dentty 319 Sσ n,2k + F X Sσ n+1,2k+1 + F X hod? Remarkaby, the answer s for a baanced poynomas F X We prove ths cam next However, ts proof depends on the foowng cassca resut Lemma 34 Lucas Theorem Let n be a natura number wth 2-adc expanson n 2 a1 + 2 a a The bnoma coeffcent n k s odd f and ony f k s ether 0 or a sum of some of the 2 a s Theorem 35 Suppose that k 1 s an nteger Consder a Booean poynoma F X n j fxed varabes Then, Sσ n+j,2k + F X Sσ n+1+j,2k+1 + F X for every postve nteger n f and ony f F X s baanced Proof The necessary part of the statement s not hard to estabsh To see ths, suppose that F X s not baanced Reca that j m m 320 Sσ n+j,2k + F X C m F S σ n,2k n j n C m F 1 m m 2k 0 j m m Sσ n+1+j,2k+1 + F X C m F S σ n+1,2k+1 n+1 j n C m F 1 m m 2k Theorem 32 mpes that {Sσ n+j,2k + F X} and {Sσ n+1+j,2k+1 + F X} satsfy the same near recurrence of order 2k + 1 It s not hard to see that the frst nta condton of {Sσ n,2k + F X} s SF, whe the frst nta condton of {Sσ n+1,2k+1 + F X} s 2SF Snce SF 0, then {Sσ n+j,2k + F X} and {Sσ n+1+j,2k+1 + F X} are dfferent sequences The suffcent part can aso be argued from the pont of vew of recurrences, but the proof becomes cumbersome as soon as j 5 Therefore, we must fnd an aternatve proof Suppose that F X s baanced For smpcty, et C m C m F and remember that SF C 0 + C C j The baancedness of F X mpes that 321 C 0 + C C j 0 We now smpfy the formuas n 320 For ths, terate the recursve formua n n 1 n to obtan the dentty 323 Thus, 324 m Sσ n+j,2k + F X Sσ n+1+j,2k+1 + F X m k n j 0 n j 0 + m k C m 1 +m 2k C m 1 +1+m 2k+1 n, n + 1

8 8 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA Observe that Sσ n+1+j,2k+1 + F X can be re-wrtten as n j [n Sσ n+1+j,2k+1 + F X C m 1 2k+1 +m 0 n j 0 The next step s to use the assumpton of baancedness of F X Snce 325 C 0 C 1 C 2 C j, then 326 Sσ n+j,2k + F X j j m1 C m C m + ] n 1 [ C m 1 2k+1 +m m 2k+1 ] n n 1 +m 2k 0 n 0 n [ 1 +m 2k 1 2k ] n and 327 Sσ n+1+j,2k+1 + F X j j m1 One consequence of Lucas Theorem s C m C m n 0 n 0 [ 1 2k+1 +m m 2k+1 ] n [ 1 2k+1 +m m 2k+1 1 2k+1 1 2k+1 ] +1 n m 2k 1 2k 1 2k+1 +m m 2k+1 1 2k+1 1 2k+1 +1 for a postve ntegers k and a non negatve ntegers and m Ths concudes the proof Exampe 36 Consder the rotaton RX X 1 X 2 + X 2 X 3 + X 3 X 4 + X 4 X 5 + X 5 X 1 Observe that RX s baanced Theorem 35 mpes that Sσ n,2k + RX Sσ n+1,2k+1 + RX for every postve nteger n and k Indeed, et 2k 10, then the frst few vaues of the sequence {Sσ n,10 + RX} startng from n 10 are 2, 24, 136, 528, 1612, 4144, 9336, 18928, 35220, 61104, , whe the frst few vaues of {Sσ n,11 + RX} startng from n 10 are Exampe 37 Consder the Booean poynoma 0, 2, 24, 136, 528, 1612, 4144, 9336, 18928, 35220, 61104, F X X 1 X 5 + X 4 X 5 + X 9 X 5 + X 12 X 5 + X 3 X 6 + X 2 X 7 + X 1 X 8 + X 1 X 9 +X 4 X 9 + X 8 X 9 + X 3 X 10 + X 7 X 10 + X 2 X 11 + X 6 X 11 + X 1 X 12 Theorem 32 mpes that the sequences {Sσ n,2 + F X} and {Sσ n+1,3 + F X} satsfy the same near recurrence of order 3 Now observe that 329 Sσ 13,2 + F X Sσ 14,3 + F X 0 Sσ 14,2 + F X Sσ 15,3 + F X 256 Sσ 15,2 + F X Sσ 16,3 + F X 512 Thus, {Sσ n,2 + F X} {Sσ n+1,3 + F X} hods for every n In vew of Theorem 35, t must be that F X s baanced Indeed, SF 0

9 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 9 It s often the case n mathematcs that resuts can be generazed by coser nspecton of ther proofs Ths s the case for Theorem 35 A cruca eement of ts proof s the dentty m 2k 1 2k 1 2k+1 +m m 2k+1 1 2k+1 1 2k+1 +1, whch s true for a postve ntegers k and a non negatve ntegers and m Now, teraton of 322 eads to the equaton 331 and other smar ones for, 2k + m, 2k m 2k t t + m t 2k + m + 1, 2k + 1 Ths and equaton 330 shfted by t mpy that + 1, and 2k + 1 2k t t 2k +m t 1 t 2k 1 t t 2k+1 +m t + 1 t 2k+1 +m+1 1 t t t 2k+1 1 t +1 2k+1 s true for a postve ntegers k and a non negatve ntegers and m Suppose that F X s a baanced Booean poynoma n the varabes X 1,, X j wth j fxed Foowng the proof of Theorem 35 and usng equaton 332 ead to the equaton [ t [ t 333 S ] t σ n+j,2k + F X S ] t σ n+1+j,2k+1 + F X whch s true for a postve ntegers k and a non-negatve ntegers t Ths eads to the foowng resut Theorem 38 Suppose that k and t are ntegers wth k postve and t non-negatve Consder a Booean poynoma F X n j fxed varabes Then, [ t ] [ t ] t t S σ n+j,2k + F X S σ n+1+j,2k+1 + F X for every postve nteger n f and ony f F X s baanced Proof Lemma 31 and Theorem 32 can be easy extended to perturbatons of the form σ n,[k1,,k s] + F X Once ths s done, the necessary part of the statement foows as n Theorem 35 The suffcent part foows from the dscusson above Exampe 39 Suppose that F X s a baanced Booean poynoma n the varabes X 1,, X j wth j fxed Theorem 38 mpes that the foowng equatons are true for every natura number n Sσ n+j,6 + F X Sσ n+j+1,7 + F X Sσ n+j,[6,5] + F X Sσ n+j+1,[7,6] + F X Sσ n+j,[6,4] + F X Sσ n+j+1,[7,5] + F X Sσ n+j,[6,5,4,3] + F X Sσ n+j+1,[7,6,5,4] + F X Sσ n+j,[6,2] + F X Sσ n+j+1,[7,3] + F X Sσ n+j,[6,5,2,1] + F X Sσ n+j+1,[7,6,3,2] + F X Sσ n+j,[6,4,2,0] + F X Sσ n+j+1,[7,5,3,1] + F X Sσ n+j,[6,5,4,3,2,1] + F X Sσ n+j+1,[7,6,5,4,3,2,1] + F X,,

10 10 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA 4 Dophantne equatons wth bnoma coeffcents In ths secton we are nterested n Dophantne equatons of the form n n 41 x 0, 0 where x beongs to some fxed bounded subset Γ of Z Exponenta sums of symmetrc Booean functons are naturay connected to ths probem Reca that exponenta sums of symmetrc Booean functons can be expressed n terms of bnoma sums To be specfc, f 1 k 1 < < k s are ntegers, then 42 Sσ n,[k1,,k s] n 0 1 k n ks Therefore, every tme we fnd a baanced symmetrc Booean functon, we aso fnd a souton to 41 where the x s beong to the set Γ { 1, 1} The converse s aso true, that s, f we fnd a souton to 41 over Γ { 1, 1}, then we aso fnd a baanced symmetrc Booean functon For exampe, consder the equaton The correspondng baanced symmetrc Booean functon s σ 8,[1,2,3,5,7] When the set consdered s Γ { 1, 1}, then any souton to 41 s sad to gve a bsecton of the bnoma coeffcents n, 0 n Observe that such souton provdes us wth two dsjonts sets A and A such that A A {0, 1, 2,, n} and n n 2 n 1 A A As mentoned n the ntroducton, the probem of bsectng bnoma coeffcents was frst dscussed by Mtche [17] Contnue wth equaton 41 over Γ { 1, 1} If n s even, then δ ± 1, for 0, 1,, n, are two soutons to 41 On the other hand, f n s odd, then the symmetry of the bnoma coeffcents mpes that δ 0,, δ n 1/2, δ n 1/2,, δ 0 are 2 n+1/2 soutons to 41 These are caed trva soutons A baanced symmetrc Booean functon n n varabes whch corresponds to one of the trva soutons of 41 over Γ { 1, 1} s sad to be a trvay baanced functon Computatons suggest that a majorty of the baanced symmetrc Booean functons are trvay baanced, thus t s of great nterest to fnd non-trvay baanced symmetrc Booean functons In the terature, these functons are caed sporadc baanced symmetrc Booean functons For exampe, the reaton 43 s not trva, therefore σ 8,[1,2,3,5,7] s a sporadc baanced symmetrc Booean functon See [9, 10, 14] for more nformaton In [21], Sarkar and Matra showed that there s an nfnte amount of sporadc baanced symmetrc Booean functons The above dscusson shows the nk between baancedness of symmetrc Booean functons and soutons to 41 over the set Γ { 1, 1} We now turn our attenton to baancedness of perturbatons of the form σ n,[k1,,k s] + F X, where 1 k 1 < < k s are ntegers and F X s a non-zero Booean poynoma n j fxed varabes, and ts connecton to equaton 41 For smpcty of the wrtng, consder the case σ n,k + F X Reca that j m m 44 Sσ n,k + F X C m F S σ n j,k, where C m F, for m 0, 1,, j, are gven by 45 C m F x F j 2 wth w2xm 1 F x Expressng exponenta sums of symmetrc Booean functons as bnoma sums eads to n j n 46 Sσ n+j,k + F X C m F 1 m m k 0

11 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 11 Observe that 47 Aso, 48 j j k C m F 1 m m C m F 1 m m k j C m F j j 2 j m j C m F SF 0 mod 2 Therefore, baancedness of a perturbaton of the form σ n+j,k + F X s connected to soutons of 41 over the set Γ e j {x 2Z : x 2 j } {0, ±2, ±4, ±6,, ±2 j } Any souton to 41 over Γ e j can be dvded by 2 to produce a souton of 41 over the set Γ j {x Z : x 2 j 1 } {0, ±1, ±2, ±3,, ±2 j 1 } The opposte s ceary true, that s, any souton to 41 over Γ j can be mutped by 2 to produce a souton over Γ e j Therefore, n ths study, most of the resuts are wrtten n the anguage of the set Γ j Remark 41 The same concuson can be reached from a perturbaton of the form σ n+j,[k1,,k s] + F X Aso, observe that f F X s the zero poynoma, then we are back at the probem of bsectng bnoma coeffcents and therefore the correspondng set s Γ { 1, 1} As n the case for bsectons of bnoma coeffcents, we can defne trva soutons to 41 over Γ j If n s odd, then the symmetry of the bnoma coeffcents mpes that 49 δ 0,, δ n 1/2, δ n 1/2,, δ 0, wth δ Γ j, are 2 j + 1 n+1 2 soutons to 41 If n s even, then 410 δ 1 m, for 0, 1,, n, and m Γ j are 2 j + 1 soutons to 41 over Γ j Aso, the symmetry of the bnoma coeffcents mpes that 411 δ 0,, δ n/2 1, 0, δ n/2 1,, δ 0, wth δ Γ j are 2 j + 1 n 2 soutons to 41 over Γ j observe that the trva souton 0, 0,, 0 s of the forms 410 and 411 Soutons of the form 49 for n odd or of the forms 410 or 411 for n even are caed trva soutons to 41 over Γ j There are other soutons whch at frst sght do not seem to be trva, for exampe, However, usng the symmetry of bnoma numbers, equaton 412 can be re-wrtten as , whch s of the form 410 Ths nvtes us to defne equvaence of soutons We say that two soutons δ 1 0, δ1 1,, δ1 n and δ 2 0, δ2 1,, δ2 n are equvaent, and wrte δ 1 0, δ1 1,, δ1 n δ 2 0, δ2 1,, δ2 n, f 1 both are non-zero soutons and 1 δ 2 0 g, δ2 1,, δ2 n ± 1 δ g, δ1 1,, δ1 n, 1 where g gcdδ 0, δ 1,, δ n, or 2 one souton can be obtaned from the other by usng the symmetry of the bnoma numbers, as t s the case of 412 and 413, or

12 12 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA 3 one souton can be obtaned from the other by combnng the prevous two cases Because of ths, we now say that soutons of the form 49, 410 or 411 are wrtten n trva form or just that they are trva form soutons and extend the defnton of trva souton to any souton that s equvaent to one of the trva form soutons For exampe, 412 s a trva souton For δ δ 0, δ 1,, δ n Γ j, defne [δ 0, δ 1,, δ n ] {δ 0, δ 1,, δ n Γ j δ 0, δ 1,, δ n δ 0, δ 1,, δ n }, that s, [δ 0, δ 1,, δ n ] s the equvaence cass of δ under Observe that f n s odd, then every trva form souton, and therefore every trva souton, beongs to the cass [0, 0,, 0] If n s even, then every trva souton beongs to ether [0, 0,, 0] or [1, 1, 1, 1,, 1, 1] As expected, the number of soutons to 41 over Γ j grows exponentay n n Ths can aready be seen n the number of trva form soutons, snce there are 2 j +1 n+1 2 such soutons for n odd and 2 j +1 n 2 +2 j for n even In fact, et Ωn, j {δ Γ n j δ s a souton to 41} and defne γ jn : Ωn, j, that s, the number of soutons to 41 that e n Γ j Tabe 1 contans γ j n for varous n s and j s Tabe 1 Number of soutons to 41 that e n Γ j n γ 1 n γ 2 n γ 3 n * * * γ 4 n * * * * * γ 5 n * * * * * * γ 6 n * * * * * * * γ 7 n * * * * * * * Remark 42 The reader s nvted to read the very nterestng paper of Ionaşcu, Martnsen and Stǎncǎ about bsectng bnoma coeffcents [13] As part of ther work, they consder the number of nontrva bsectons They found many prevousy unknown nfnte casses of ntegers whch admt nontrva bsectons, and a cass wth ony trva bsectons They aso found severa bounds for the number of nontrva bsectons and compute the exact number of such bsectons for n 51 Some of these resuts may be extended to perturbatons of symmetrc Booean functons For exampe, foowng smar technques form [13] we obtaned an ntegra representaton for γ j n Let V j [0, 2 j 1 ] Z The notaton x x 0, x 1,, x n s used to represent a vector x V n+1 j For x V n+1 j, et wx represents the number of non-zero entres of x The number γ j n s gven by the foowng ntegra 414 γ j n x V n+1 j 2 wx 1 0 n n cos πx s ds Many of the soutons that are counted n Tabe 1 are equvaent to some others Therefore, the amount of meanngfu soutons shoud be sgnfcanty smaer than the numbers that are presented n Tabe 1 Defne ω j n to be the number of dfferent equvaence casses on Ωn, j under, that s, the cardnaty of the quotent set Ωn, j/ For exampe, 415 Ω4, 2/ {[0, 0, 0, 0, 0], [0, 1, 2, 2, 0], [2, 2, 1, 0, 0], [2, 1, 1, 0, 0], [2, 1, 0, 0, 2]}, and therefore ω Tabe 2 contans ω j n for varous n s and j s The order of magntude of these numbers s smaer than the ones on Tabe 1, as expected, however, t s apparent that there are many meanngfu soutons that are not trva A baanced perturbaton of the form σ n+j,[k1,,k s] + F X s sad to be trvay baanced f t corresponds to one of the trva soutons of 41 over Γ j Anaogous to the case for bsectons of bnoma coeffcents, a non-trvay baanced perturbaton of the form σ n+j,[k1,,k s] + F X s caed sporadc baanced Booean perturbaton As t s the case of baanced symmetrc Booean functons, many baanced perturbatons seem to be trvay baanced For exampe, consder σ n,k +X 1, whch s the smpest perturbaton to an eementary symmetrc Booean poynoma We have the foowng resut

13 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 13 Tabe 2 Number of dfferent equvaence casses on Ωn, j under n ω 1 n ω 2 n ω 3 n ω 4 n * ω 5 n * * * ω 6 n * * * * ω 7 n * * * * * Theorem 43 Let k be a postve nteger Let r og 2 k + 1 The perturbaton σ n,k + X 1 s trvay baanced for n 2 r m + k 1 where m s any natura number Proof Reca that 416 Sσ n,k + X 1 Sσ n 1,k Sσ n 1,[k,k 1] n 1 1 n 1 k n 1 1 k+ 0 0 k 1 n 1 Let Nk be the set of ntegers such that k s odd Smary, defne Nk, k 1 to be the set of ntegers such that k + k 1 s odd Lucas Theorem mpes that Nk f and ony f 1 Nk, k 1 Aso, the choce n 2 r m + k 1 mpes that f 0 n, then Nk f and ony f n Nk Suppose that 0 Nk, that s, that the coeffcent of n n the frst sum of 416 s 1 By the choce of n one has n 0 Nk But f ths s true, then n 0 1 Nk, k 1 and therefore the coeffcent of n 1 n 1 n n 0 1 n 1 n 0 1 n the second sum of 416 s aso 1 The converse s aso true If 1 Nk, k 1, then the coeffcent of n n the second sum of 416 s 1 But then, Nk and by the choce of n one has that n n 1 1 Nk Thus, the coeffcent of n 1 n n n the frst sum of 416 s aso 1 We concude that σ n,k + X 1 s trvay baanced Exampe 44 Consder the perturbaton σ 20,5 + X 1 Theorem 43 mpes that ths perturbaton s trvay baanced Expcty, 19 Sσ 20,5 + X trvay We had tred to fnd sporadc baanced perturbatons of the form σ n,k + X 1, but our attempts faed Ths ed us to beeve that a conjecture smar to the one presented by Cusck, L and Stǎncǎ n [10] for the case of eementary symmetrc Booean functons hods aso see [11] To be specfc, we conjecture the foowng: 0

14 14 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA Conjecture 45 No perturbaton of the form σ n,k + X 1 s baanced except for the trva cases, e when n 2 r m + k 1 where r og 2 k s + 1 and m s a postve nteger Smar technques can be used to show that other nfnte fames are trvay baanced However, we w not show them here, except for the next exampe for whch we omt the proof Theorem 46 The perturbaton 422 σ 2 +1 D 1,2 + X 1 + X X 2m, where D, and m are postve ntegers, s trvay baanced In vew of Theorem 35, the perturbaton 423 σ 2 +1 D, X 1 + X X 2m s aso trvay baanced As t s the case of symmetrc Booean functons, computatons suggest that trvay baanced perturbatons are qute common Therefore, t s a good dea for us to study them n more deta Observe that Theorem 43 mpes that once σ n,k + X 1 s trvay baanced at one pont, for exampe, when n 2 r + k 1, then σ n,k + X 1 s trvay baanced for nfntey many n, e for n 2 r m + k 1 where m s a postve nteger Ths s not a concdence, n fact, t s true for every perturbaton Let 1 k 1 < < k s be ntegers Consder σ n+j,[k1,,k s] + F X where F X s a Booean poynoma n the varabes X 1,, X j j fxed Suppose that σ n+j,[k1,,k s] + F X s trvay baanced, that s, t corresponds to a trva souton δ 0, δ 1,, δ n to 41 over Γ j For the sake of smpcty, suppose that δ 0, δ 1,, δ n s a trva souton of the form δ δ n The vaue of δ s gven by 424 and we know that 425 δ j j m C m F 1 m [ k ks ] C m F 1 +m k m ks, + m k 1 + m,, are a perodc moduo 2 wth a perod ength 2 r where r og 2 k s + 1 Ths means that but then, In other words, the tupe k s δ δ +2 r and δ n δ n +2 r, δ δ n δ n+2r δ 0, δ 1,, δ n+2 r s a trva souton to 41 over Γ j However, ths tupe corresponds to Sσ n+2r +j,[k 1,,k s] + F X, e n+2 r n + 2 r 426 Sσ n+2 r +j,[k 1,,k s] + F X 2δ 0 Ths dscusson eads to the foowng resut Theorem 47 Let 1 k 1 < < k s be ntegers and F X be a Booean poynoma n the varabes X 1,, X j j fxed Let r og 2 k s + 1 Suppose that n 0 s a postve nteger such that 427 σ n0+j,[k 1,,k s] + F X s trvay baanced Then, 428 σ n0+m 2 r +j,[k 1,,k s] + F X s trvay baanced for every non-negatve nteger m Proof Ths s a drect consequence of the above dscusson 0

15 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 15 Exampe 48 Consder the perturbaton σ 23,8 + F X where F X X 1 + X 2 + X 3 + X 4 + X 5 + X 6 + X 7 + X 8 + X 9 Ths perturbaton s actuay baanced It corresponds to the equaton δ 0, where δ δ 0, δ 1,, δ , 28, 56, 70, 56, 28, 8, 0, 8, 28, 56, 70, 56, 28, 8 Therefore, σ 23,8 +F X s trvay baanced Theorem 47 mpes that σ 23+16m,8 +F X s trvay baanced for every non-negatve nteger m Now, snce F X s baanced and σ 23+16m,8 + F X s baanced, then Theorem 35 mpes that the perturbaton σ 24,9 + F X s aso baanced In fact, the perturbaton s trvay baanced as expected and t corresponds to the equaton δ 0, where δ δ 0, δ 1,, δ , 9, 37, 93, 163, 219, 247, 255, 255, 247, 219, 163, 93, 37, 9, 1 Theorem 47 mpes that σ 24+16m,9 + F X s trvay baanced for every non-negatve nteger m Exampe 49 Consder the perturbaton σ 7,4 + X 1 + X 2, whch s trvay baanced The souton to 41 over Γ 2 that corresponds to Sσ 7,4 + X 1 + X 2 s In vew of Theorem 47, we concude that σ 7+8m,4 + X 1 + X 2 s trvay baanced for every non-negatve nteger m Now appy Theorem 35 to concude that σ 8+8m,5 + X 1 + X 2 s baanced for every non-negatve nteger m Observe that the souton to 41 over Γ 2 that corresponds to Sσ 8,5 + X 1 + X 2 s , whch, at frst, does not ook ke a trva souton unt we reaze that t s equvaent to Thus, σ 8,5 + X 1 + X 2 s trvay baanced The same concuson can be reached for σ 8+8m,5 + X 1 + X 2 Exampe 410 The reader can check that σ 15,[3,4,5,6] +X 1 +X 2 +X 3 s baanced and that ts correspondng souton to 41 over Γ 3 s δ δ 0, δ 1,, δ 12 1, 2, 0, 2, 1, 0, 0, 0, 1, 2, 0, 2, 1 Therefore, σ 15+8m,[3,4,5,6] + X 1 + X 2 + X 3 s trvay baanced for every non-negatve nteger m Ths famy of functons can be fted up by Theorem 38 to concude that σ 16+8m,[4,5,6,7] + X 1 + X 2 + X 3 s aso baanced for every non-negatve nteger m Let us recap what we have so far Exponenta sums of perturbatons of symmetrc Booean functons are connected to soutons of the equaton 41 over some bounded subsets of the ntegers As t s the case for symmetrc Booean functons, the concept of trva and non-trva soutons can be defned as we as the concepts of trvay baanced perturbaton and sporadc baanced perturbaton Smar to the case of symmetrc Booean functons, computatons suggest that many baanced perturbatons turn out to be trvay baanced Moreover, Theorem 47 states that once a perturbaton of fxed degree s trvay baanced at one n, then t s trvay baanced for nfntey many n

16 16 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA Wth regard to ths ast theorem, n the case of a symmetrc Booean functon of fxed degree, ony trvay baanced cases exst when the number of varabes grows Ths was frst conjectured by Canteaut and Vdeau [4] and proved by Guo, Gao and Zhao [12] In the next secton, we estabsh a smar resut for the type of perturbatons consdered n ths artce The technques to be used are nspred by the ones presented n [12] 5 Baancedness of perturbatons as the number of varabes grows In ths secton we consder baancedness of perturbatons when the number of varabes grows Ths probem provdes nformaton about the converse of the statement of Theorem 47 The technques used n ths secton are very smar to the ones presented n [12] Consder a perturbaton of the form σ n,[k1,,k s] + F X where 1 k 1 < < k s are ntegers and F X s a poynoma n the varabes X 1,, X j wth j fxed The probem s to characterze when Sσ n,[k1,,k s] + F X 0 for n bg enough Reca that the exponenta sum Sσ n,[k1,,k s] + F X satsfes recurrence 26, that s, t satsfes 2 r x n 1 1 r x n, 1 where r og 2 k s + 1 In other words, the sequence of exponenta sums of our perturbaton s a rea souton to the near recurrence 51 Therefore, we frst answer the queston of when a rea souton {a n } to 51 s zero for some n bg enough The characterstc poynoma assocated to 51 s 52 X 2Φ 4 X 1Φ 8 X 1 Φ 2 rx 1 Ths mpes that any souton {a n } to 51 has the form 2 r 1 53 a n d 0 2 n + d λ n, where λ 1 + ξ 1 1 wth ξ exp π 2 r 1 and 1 Observe that ξ2r ξ, λ 2r λ, and λ 2 r 1 0 Moreover, f {a n } s a rea souton, then we aso have d 2r d From now on, suppose that {a n } s a rea souton to 51 We now foow [12] and express {a n } as 2 r a n d 0 2 n + 2 Red λ n, where Rez denotes the rea part of z, and defne 55 t n Red λ n, for 0 2 r 1 1 Ths eads to 2 r a n t 0 n + 2 t n The next emma gves a characterzaton of when a n equas zero for some n bg enough It s bascay the same statement as Lemma 3 of [12], whch tsef s a modfcaton of a proof of Ca, Green and Therauf [3] The proof foows amost verbatm as the one n [12] Lemma 51 Suppose that {a n } s a rea souton to the near recurrence 51 Then there exsts an nteger n 0 such that for any n > n 0, 1 1 a n 0 f and ony t n 0, for 0 2 r 1 1 The number t n can be expressed as n π 57 t n d 2 cos 2 r cos argd πn 2 r

17 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 17 Ths mpes that Remark 2 of [12] aso appes and therefore, t n 0 f and ony f d ξ n d, for any 0 2 r 1 1 Ths characterzes the cases when a rea souton a n to recurrence 51 s zero for n bg, that s, there exsts an nteger n 0 such that for any n > n 0, a n 0 f and ony d ξ n d, for any 0 2 r 1 1 Let us go back to our perturbatons Agan, for smpcty, consder a perturbaton of the form σ n,k +F X where F X s a Booean poynoma n the varabes X 1,, X j wth j fxed Reca that n j n 58 Sσ n+j,k + F X C m F 1 m m k Let δ F k be defned as 59 δ F k In other words, 58 can be re-wrtten as 0 j 510 Sσ n+j,k + F X C m F 1 m m k n 0 δ F k Observe that f we fnd a vaue of n for whch 58 s baanced, then we fnd a souton to the Dophantne equaton n n 511 x 0, 0 over Γ e j and the souton woud be gven by δ F 0 k, δ F 2 k,, δ n F k Reca that every souton to 51 has the form 53 The exponenta sum of our perturbatons, as we as the ones of symmetrc Booean functons, satsfy 51, therefore they can be expressed n form 53 In the case of the symmetrc Booean functon σ n,[k1,,k s], where 1 k 1 < < k s are ntegers, Ca, Green and Therauf [3] found an expct formua for the coeffcents n 53, that s, f r og 2 k s + 1, then 2 r Sσ n,[k1,,k s] c 0 k 1,, k s 2 n + c k 1,, k s λ, where 1 n 513 c k 1,, k s 1 2 r 1 2 r 1 k ks ξ In vew of equaton 213, ths mpes that, for r og 2 k + 1, one has 2 r Sσ n+j,k + F X d 0 2 n + d λ n, where 1 j 2 1 r 1 d C m F 2 r 1 m m k a ξ a a0 1 2 r 1 j 2 r C m F 1 m m k a a0 1 2 r 1 2 r δ a F k ξ a a0 ξ a

18 18 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA Remark 52 Ths dscusson carres over to perturbatons of the form σ n,[k1,,k s] + F X wthout too much effort Wth ths nformaton at hand, we are ready to present one of the man resuts of ths artce Ths resut s a generazaton of Canteaut and Vdeau s observaton for symmetrc Booean functons of fxed degree To be specfc, we show that, excudng the trva cases, baanced perturbatons of fxed degree do not exst when the number of varabes grows Theorem 53 Suppose that 1 k 1 < < k s are ntegers and F X s a Booean poynoma n the varabes X 1,, X j j fxed There exsts an n 0 such that for every n > n 0, σ n+j,[k1,,k s] + F X s baanced f and ony f t s trvay baanced Proof We present the proof for the case of a perturbaton of the form σ n+j,k + F X The genera case foows by the same argument Reca that {Sσ n+j,k + F X} s a rea souton to 51 Therefore, there exsts an nteger n 0 such that for any n > n 0, 515 Sσ n+j,k + F X 0 f and ony f d ξ n d, for any 0 2 r 1 1, where 516 d 1 2 r 1 j 2 r ξ n d ξn 2 r a0 1 2 r 1 2 r δ a F k ξ a, a0 a0 C m F 1 m m a k and r og 2 k + 1 Suppose that n > n 0 and that d ξ nd for any 0 2 r 1 1 Observe that 2 r 1 j C m F 1 m 517 m k a 1 2 r 1 j 2 r a0 1 2 r n tn 2 r r 1 j 2 r a0 C m F 1 m m a k j 1 2 r 1 2 r δ n ak F ξ a, a0 ξ a ξ a ξ n a C m F 1 m m n t k C m F 1 m m n a k j where the prevous to the ast dentty hods because C mf 1 m m n a k ξ a has perod 2 r Therefore, d ξ n d for any 0 2 r 1 1, hods f and ony f r 1 δ a F k + δ n akξ F a 0, for any 0 2 r 1 1 a0 Let fx be the poynoma 2 r fx δ a F k + δ n akx F a a0 ξ a ξ t

19 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 19 Observe that equaton 518 mpes that a poynomas on the st X 1, X 2 + 1, X 4 + 1,, X 2r dvde fx But a these poynomas are rreducbe n Q[X], therefore r X 1 X 2t + 1 dvdes fx t1 However, the degree of both X 1 r 1 t1 X2t +1 and fx s 2 r 1 Snce Q[X] s a Unque Factorzaton Doman, then t foows that r F X z X 1 X 2t + 1, for some constant z t s not hard to see that z must be an nteger But r X 1 X 2t X X 2 + X 3 X 4 + X 2r 2 + X 2r 1 t1 Equatons 519, 521 and 522 yed 523 δ F a k + δ F n ak 1 a 1 z for 0 a 2 r 1 Note that equaton 523 hods beyond the range 0 a 2 r 1 because δ F a k has perod 2 r Thus, when n s bg enough, equaton 523 characterzes a soutons t1 δ F 0 k, δ F 1 k, δ F 2 k,, δ n F k to the Dophantne equaton 511 over the set Γ e j that come from our perturbaton The next step s to show that a of them are trva Suppose frst that n s odd, that s, suppose that n 2m + 1 We know that 524 δ F a k + δ F 2m+1 a k 1a 1 z, where z s a fxed nteger Let a m, then 525 δ F m k + δ F m+1 k 1m 1 z On the other hand, et a m + 1 Then, 526 δ F m+1 k + δf m k 1 m z Equatons 525 and 526 mpy that z 0 But then 527 δ n ak F δ a F k and we concude that the perturbaton s trvay baanced when n s odd Suppose now that n s even, e n 2m As before, we know that 528 δ F a k + δ F 2m a k 1a 1 z, where z s a fxed nteger If z 0, then t s cear that the perturbaton s trvay baanced Thus, suppose that z 0 Let a m n equaton 528 Then, 529 2δ F m 1 m 1 z and therefore z s even Say z 2z 0 wth z 0 a non-zero nteger Then, δ F m δ F 0 k, δ F 1 k,, δ F 2m k m 1 z 0 and δ F 0 k + δ F 2m k, δf 1 k + δ F 2m 1 k,, δf m 1 k + δf m+1 k, δf m k, 0, 0,, 0 2z 0, 2z 0,, 1 m 2z 0, 1 m 1 z 0, 0, 0,, 0 2, 2,, 1 m 2, 1 m 1, 0, 0,, 0 1, 1, 1, 1,, 1, 1

20 20 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA We concude that the perturbaton s aso trvay baanced when n s even Snce t s cear that trvay baanced mpes baanced, then we concude that there s an nteger n 0 such that for any n > n 0, the perturbaton σ n,k + F X s baanced f and ony f t s trvay baanced Ths concudes the proof 6 Some exampes of sporadc baanced perturbatons In the prevous secton we earned that, excudng the trva cases, baanced perturbatons of fxed degree do not exst when the number of varabes grows Thus, as n the case for symmetrc Booean functons, t s of great nterest to fnd sporadc baanced perturbatons Next are some exampes of sporadc baanced perturbatons and ther correspondng soutons to 41 Exampe 61 The perturbaton σ 24,14 + X 1 + X 2 s baanced and sporadc The souton to 41 over Γ 2 that corresponds to Sσ 24,14 + X 1 + X 2 s Observe that Theorem 35 mpes that σ 25,15 + X 1 + X 2 s aso baanced and sporadc The souton to 41 over Γ 2 that corresponds to Sσ 25,15 + X 1 + X 2 s Equaton 62 corresponds to one souton to the foowng three consecutve bnoma coeffcents Dophantne equaton n n n k + 2 k + 1 k Luca and Szaay [16] studed equatons of the form n n 64 A + B k k + 1 n + C 0, k + 2 where A, B, C Z, A > 0, C 0 and gcda, B, C 1 As part of ther study, they provded nfntey many soutons to 63 gven by t 2 2 t 2 2 t t t 4/2 t t 2/2 t 2 0, + t/2 where t s any nteger satsfyng t 3 Observe that 62 corresponds to 65 when t 5 Moreover, 61 can be obtaned from 62 by appyng the dentty n n 1 n k k k 1 Exampe 62 In [23], Sngmaster shows that the Dophantne equaton n n n 67 +, k k + 1 k + 2 has nfntey many soutons gven by n F 2+2 F and k F 2 F 2+3 1, where F n represent the n-th Fbonacc number The smaest of these soutons s gven by F4 F 5 1 F4 F 5 1 F4 F F 2 F 5 1 F 2 F 5 F 2 F Observe 68 s a non-trva souton to 41 over Γ 1 A nce probem s to fnd sporadc baanced perturbatons of the form σ 15,[k1,,k s] + X 1 that corresponds to 68 There are ony four of such perturbatons

21 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 21 of degree ess than or equa to 14 The frst one s σ 15,[5,6,10,12,13] + X 1, whch corresponds to the equvaent souton Another s σ 15,[6,8,9,10,13] + X 1, whch corresponds to the equvaent souton The thrd one s σ 15,[6,7,11,12,14] + X 1 and t corresponds to the equvaent souton Fnay, the ast one s σ 15,[5,6,7,8,9,11,14] + X 1 The correspondng equvaent souton s Exampe 63 Let us go back to the equaton 65 Note that the smaest souton to ths equaton s gven by Tabe 3 ncudes a sporadc baanced perturbatons of the form σ 8,[k1,,k s] + X 1 for k s 7 for whch ther correspondng soutons to the Dophantne equaton 41 are equvaent to 613 Tabe 3 Perturbatons and ther correspondng soutons to 41 Perturbaton Correspondng souton σ 8,[3,6] + X 1 0, 0, 1, 1, 0, 1, 1, 0 σ 8,[1,2,6] + X 1 1, 0, 1, 0, 1, 1, 1, 1 σ 8,[1,5,6] + X 1 1, 1, 1, 1, 0, 1, 0, 1 σ 8,[2,3,5,6] + X 1 0, 1, 1, 0, 1, 1, 0, 0 σ 8,[1,4,7] + X 1 1, 1, 1, 0, 1, 1, 0, 1 σ 8,[2,3,4,7] + X 1 0, 1, 1, 1, 0, 1, 0, 0 σ 8,[3,4,5,7] + X 1 0, 0, 1, 0, 1, 1, 1, 0 σ 8,[1,2,4,5,7] + X 1 1, 0, 1, 1, 0, 1, 1, 1 Tabe 4 ncudes a sporadc baanced perturbatons of the form σ 9,[k1,,k s] + X 1 + X 2 for k s 8 for whch ther correspondng soutons to the Dophantne equaton 41 are equvaent to 613 Smar exampes can be produced wth the ad of computers A Mathematca mpementaton can be found n For exampe, usng ths mpementaton, we found that there are 265 sporadc baanced perturbatons of the form σ n,[k1,,k s] + X 1 wth n, k s 17 Aso, there are 606 sporadc baanced perturbatons of the form σ n,[k1,,k s] + X 1 + X 2 wth n, k s 17 The reader s nvted to use ths Mathematca mpementaton to fnd more Acknowedgments The second author was partay supported as a student by NSF-DUE and the Meon-Mays Undergraduate Feowshp

22 22 FRANCIS N CASTRO, OSCAR E GONZÁLEZ, AND LUIS A MEDINA Tabe 4 Perturbatons and ther correspondng soutons to 41 Perturbaton Correspondng souton σ 9,[3,6] + X 1 + X 2 0, 0, 1, 1, 0, 1, 1, 0 σ 9,[3,7] + X 1 + X 2 0, 1, 2, 1, 0, 0, 0, 0 σ 9,[6,7] + X 1 + X 2 0, 0, 0, 0, 1, 2, 1, 0 σ 9,[1,3,7] + X 1 + X 2 2, 1, 0, 1, 2, 2, 2, 2 σ 9,[1,4,7] + X 1 + X 2 2, 2, 1, 1, 2, 1, 1, 2 σ 9,[1,6,7] + X 1 + X 2 2, 2, 2, 2, 1, 0, 1, 2 σ 9,[1,2,3,7] + X 1 + X 2 1, 0, 1, 2, 1, 1, 1, 1 σ 9,[1,2,4,7] + X 1 + X 2 1, 1, 2, 2, 1, 0, 0, 1 σ 9,[1,2,6,7] + X 1 + X 2 1, 1, 1, 1, 2, 1, 0, 1 σ 9,[1,3,4,6,7] + X 1 + X 2 2, 1, 1, 2, 1, 1, 2, 2 σ 9,[1,2,3,4,6,7] + X 1 + X 2 1, 0, 0, 1, 2, 2, 1, 1 σ 9,[5,8] + X 1 + X 2 0, 0, 0, 1, 2, 2, 1, 0 σ 9,[2,5,8] + X 1 + X 2 1, 1, 1, 2, 1, 1, 2, 1 σ 9,[3,4,5,8] + X 1 + X 2 0, 1, 1, 1, 2, 1, 0, 0 σ 9,[3,5,6,8] + X 1 + X 2 0, 1, 2, 2, 1, 0, 0, 0 σ 9,[4,5,6,8] + X 1 + X 2 0, 0, 1, 2, 1, 1, 1, 0 σ 9,[2,3,4,5,8] + X 1 + X 2 1, 2, 2, 2, 1, 0, 1, 1 σ 9,[2,3,5,6,8] + X 1 + X 2 1, 2, 1, 1, 2, 1, 1, 1 σ 9,[2,4,5,6,8] + X 1 + X 2 1, 1, 0, 1, 2, 2, 2, 1 References [1] A Adophson and S Sperber p-adc Estmates for Exponenta Sums and the of Chevaey-Warnng Ann Sc Ec Norm Super, 4 e sére, , [2] J Ax Zeros of poynomas over fnte feds Amer J Math , [3] J Ca, F Green and T Therauf On the correaton of symmetrc functons Math Systems Theory , [4] A Canteaut and M Vdeau Symmetrc Booean Functons IEEE Trans Inf Theory 51, no , [5] F N Castro, O E Gonzáez and L A Medna A dvsbty approach to the open boundary cases of Cusck-L-Stǎncǎ s conjecture Cryptography and Communcatons , [6] F N Castro and L A Medna Lnear Recurrences and Asymptotc Behavor of Exponenta Sums of Symmetrc Booean Functons Eec J Combnatorcs , #P8 [7] F N Castro and L A Medna Asymptotc Behavor of Perturbatons of Symmetrc Functons Annas of Combnatorcs , [8] F N Castro and L A Medna Moduar perodcty of exponenta sums of symmetrc Booean functons Dscrete App Math , [9] T W Cusck, Y L k-th order symmetrc SAC Booean functons and bsectng bnoma coeffcents Dscrete App Math , [10] T W Cusck, Y L, and P Stǎncǎ Baanced Symmetrc Functons over GF p IEEE Trans on Informaton Theory , [11] T W Cusck, Y L, and P Stǎncǎ On a conjecture for baanced symmetrc Booean functons J Math Crypt , 1 18 [12] Y Guo, G Gao, Y Zhao Recent Resuts on Baanced Symmetrc Booean Functons IEEE Trans Inf Theory 62, no , [13] E J Ionaşcu, Thor Martnsen, Pantemon Stǎncǎ Bsectng bnoma coeffcents arxv: [mathco] [14] N Jefferes Sporadc parttons of bnoma coeffcents Eectr Lett , [15] M Koountzaks, R J Lpton, E Markaks, A Metha and N K Vshno On the Fourer Spectrum of Symmetrc Booean Functons Combnatorca ,

23 DIOPHANTINE EQUATIONS AND PERTURBATIONS OF SYMMETRIC BOOLEAN FUNCTIONS 23 [16] F Luca and L Szaay Lnear dophantne equatons wth three consecutve bonoma coeffcents Acta Academae Paedagogcae Agrenss, Secto Mathematcae , [17] C Mtche Enumeratng Booean functons of cryptographc sgnfcance J Cryptoogy 2, no , [18] O Moreno and C J Moreno Improvement of the Chevaey-Warnng and the Ax-Katz theorems Amer J Math , [19] O Moreno and C J Moreno The MacWams-Soane Conjecture on the Tghtness of the Cartz- Uchyama Bound and the Weghts of Dua of BCH Codes IEEE Trans Inform Theory , [20] O Moreno, K Shum, F N Castro and PV Kumar Tght Bounds for Chevaey-Warnng-Ax Type Estmates, wth Improved Appcatons Proc of the London Mathematca Socety , [21] P Sarkar and S Matra Baancedness and correaton mmunty of symmetrc Booean functons Dscrete Math , [22] A Shpka and A Ta On the Mnma Fourer Degree of Symmetrc Booean Functons Combnatorca , [23] D Sngmaster Repeated bnoma coeffcents and Fbonacc numbers Fbonacc Quartery , Department of Mathematcs, Unversty of Puerto Rco, San Juan, PR E-ma address: francscastr@gmacom Department of Mathematcs, Unversty of Puerto Rco, San Juan, PR E-ma address: oscargonzaez3@upredu Department of Mathematcs, Unversty of Puerto Rco, San Juan, PR E-ma address: usmedna17@upredu