ON THE DETERMINATION OF NUMBERS BY THEIR SUMS OF A FIXED ORDER J. L. SELFRIDGE AND E. G. STRAUS

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1 ON THE DETERMINATION OF NUMBERS BY THEIR SUMS OF A FIXED ORDER J. L. SELFRIDGE AND E. G. STRAUS 1. Introduction. We wih to treat the following problem (uggeted by a problem of L. Moer [2]): Let {x} = {x u,# } be a et of complex number (if one i intereted in generality, one may conider them element of an algebraically cloed field of characteritic zero) and let {σ} = {σ lf, σ, n Λ be \) the et of um of ditinct element of {x}. To what extent i {x} determined by {σ} and what et can be {σ} et? In 2 we anwer thi quetion for = 2. In 3 we treat the quetion for general. 2. The cae = 2. THEOREM 1. If n Φ2 k then the firt n elementary ymmetric function of {σ} can be precribed arbitrarily and they determine {x} uniquely. Proof. Intead of the elementary ymmetric function we conider the um of power, etting Σfc Σ σ ί * Then (2) (l) Σ* = i>?= Σ (χ h + χ H r = \ Σ Σ K + Xi/ - Σ (2a? 4 )*). Expanding the binomial and collecting like power we obtain = l A2n - 2«)S k + I Thu, ince the coefficient of S^ doe not vanih, we can olve re- Received May 16,

2 848 J. L. SELFRIDGE AND E. G. STRAUS curively for S k in term of Σi> > Σ* I* 1 particular Σi> > Σn determine S u, Sn and hence x L,, x n uniquely. THEOREM 2. If n 2 k then Σi> " > Σ&+i wwί atify a certain algebraic equation and {σ} will not alway determine {x}. Proof. Equation (1) for Σ fc+1 yield / Q \ v-i J. ^-i I rc "T" 1 An C (2) Σ.«= - 2 Σ( z jsλ +1., where S l9, S fc are expreed by (1) a polynomial in Σi> * φ t Σfc To prove the econd part of the theorem we proceed by induction. Aume there are two different et {x lf, x 2k -i}, {Vi, *,:ί/ a *-i}- which have the ame {σ}. Then conider the two et {X} = [χ λ + α,, a^-! + α, 2/i,, 2/ 2 Λ-I} = {x i9, Λj.fc-i, y\λ~ a,, 2/.fc-i 4~ <^} Clearly every um of two element of {X} i either σ t or σ 4 + 2α or a?< + 2/j + α and the ame hold for the um of two element of {Y}. The et {X}, {F} will clearly be different for ome α. To how that they are different for any a Φ 0, rearrange {x} and {y} o that α? 4 = 2/i * = 1, 2,, m m > 0, and #, ^= ι/ fc for j, k > m. Then ince 2/ 4 + α = a? t + α i = 1, 2,, m, the et {X} and {Y} will be different if {xj \j>m} i different from {x 3 + a \ j > m}. But thi i clear for any α Φ 0. Since {σ} clearly doe not determine [x] for n = 2 the proof i complete. In a ene we have completed the anwer of the quetion raied in the introduction for = 2, however there remain ome unanwered quetion in cae τι = 2\ 1. 7/ {σ} doe not determine {x} can there be more than two et giving rie to ame {σ}? The anwer i trivially " ye " for k 0,1 and i " no?> for k = 2. It eem probable that the anwer i "no " for all & > 2, however we can ee no imple way of proving thi. 2. For what value of n doe there exit for all {real) {x} a tranformation y L A (x lt, x n ), different from a permutation, o that {x} and {y} give rie to the ame {σ}? Thi quetion wa uggeted by T. S. Motzkin who gave the anwer for = 2.

3 ON THE DETERMINATION OF NUMBERS 849 LEMMA 1. // n > and the above function f i exit then they are linear. Proof. The et {y}, {x} are connected by a ytem of equation Vi L + + Vi β = x h + + x, t. Here the indice i l9 «,i are themelve function of {x}. However, ince they aume only a finite et of value, there exit a omewhere dene et of {x} for which the indice are contant. We retrict our attention to that et. Let Δ^y % =f.(x l9, x k + h,, x n ) -fi(x u, #fc>,#») then we obtain ( 3 ) Δ^y h + + 4X - 0 or A. If we let Ui be the difference of 4 Λ) 2/«for two different et of value of {x} then, ince the right-hand ide of (3) i independent of the choice of {x}, we obtain ( 4 ) u h +. + u i = 0. Summation over all et {i l9, i } c {1,, rc} yield ( 5 ) % t + u. z + + u n = 0. Now let be the leat poitive integer o that u iχ + + n %% 0 for all {ij,, ij c {1,, n}. Then ί n, for w = mi + r with 0 < r < ί implie w h + + n %r = u x + u u n - Yiμ h + + u 3 ) = 0 for all {i lf, i r } c {1,, n}, contrary to hypothei. Since n > > t we mut have n> 2b. If t > 1 then Uj= (u h + + u itmmi ) for every j $ {i u, i t -i}. But there are more than t uch j, ay j u,^t. Hence M Jt + * + U h = - ί(% tl + + M* tβl ) = 0 or u h + + ^ί_ 1 = 0 for every {i l9, i t -i} c {1,, w} contrary to hypothei. Thu = 1 and ^ u^ = ^w = 0. In other word Δ^y^akV = cont. Thu 4 Λ i>2/ι + J^ v ^ = ^'^Vi that a (^ = a ik h and o 2/ι = Σ ^i^fc

4 850 j. L, SELFRIDCE AND. C. STRAUS THEOREM 3. If n > and there exit a nontrivial tranformation Vι = fifaif, x n ) which preerve {σ} then n = 2 and the tranformation i linear with matrix (up to permutation) Proof. We know by Lemma 1 that the tranformation mut be linear. Let y i = ^ka ik x k then ( 6 ) y h + + y lg = Σ K* + Hence, for fixed fe, we have + = ^ + + (7) o for et {i if l for (gl J) et {i lf...,i β }. Since w > two element α ifc, α jfc in the ame column atify a ίk + α <ljfc + + a ι& _ ik = 0 or 1 a jk + a hk + + α ig _ ifc = 0 or 1 where {i u,*,-,] e {1, Hence } {ΐ, j}. (8) α, = Q>)k o r = a jk ± Let the two value aumed by term in the fcth column be a k and 1 + α fc. From (6) we ee that both value mut occur. On the other hand if both a k and 1 + a k would occur more than once then max(αi ljb + + a t k ) min(d lfc + + a ik ) > 2 in contradiction to (7). If 1 + a k i aumed only once, ay a kk = 1 + a Λ, then 0 = a k or (9) (10) According to (6) we have θ i Φ k. We now repeat the argument that led to equation (8). Since n >

5 we can write for any pair (i, j) ON THE DETERMINATION OF NUMBERS 851 Σ (<V + + a*,.*) + Σ α ijfc = Σ (<V + + α,,) + Σ <*>,-* = where {i^, i - τ } c {1,, n) - {i, j}. Hence ΣLi a ijb = Σ*-i ^» and according to (10), Σϊ-i a M = ^ o that (11) Σα< fc = 1 i = 1,, w. Combining (9) and (11) we obtain (12) a kj = ί 1 ^' = * 10 i = fc. In other word, every column contain 0 and therefore a k 0 for k = 1,,?2. Thu the tranformation i a permutation. The only nontrivial cae arie therefore if the value a k occur only once, ay a kk = α fc. Then 1 + a k = 0 and Combining (11) and (13) we obtain /-,,x Λ ^i n(n 1) 1 w / ^ fc = l i=l S S S and hence n = 2. It i now clear from (11) that each row and column contain exactly one term ( l)/ and that the matrix (up to permutation) i the one given in the theorem. 3. General. The procedure which led to Theorem 1 can be generalized. Firt we define, for every, a function which i a polynomial in n, 2 k, 3 fc,, fc. Let (15) j\n, k) - A Σ(- irv- 1 ΣV S P i Ί where the outer ummation i over all permutation P on β mark, each permutation being compoed of a % i-cycle i = 1, *,r, and ί = tti + + a r. Thu (16) /(n, fe) = n («- l)(2 fc + - 2)w- a + ( - l)(β - 2)[A( 3 * + -3) + (β - 3)(2 fc )~L S (- l) ( - i; o J - ( - l) ( - 1)! β*" 1.

6 852 j. L. SELFRIDGE AND E. G. STRAUS THEOREM 4. For every conider the ytem of Diophantine equation f(n, k) = 0 k = 1, 2,, n. If n atifie none of thee then the firt n elementary ymmetric function of {σ} can be precribed arbitrarily and they determine {x} uniquely. If f(n, k) = 0, then the firt k elementary ymmetric function of {σ} mut atify an algebraic equation. Proof. In the notation of Theorem 1 we have (17) Σ* = Σ K + *«,+ + χ h f =. Σ K + + *«,)* where by D(t) i meant ummation over all et of ubcript i 5 at leat t of which are ditinct. Hence! Σ* = Σ K + + a?, )* - ( ) Σ (2a? 4l + α? <a + + a?, f x ) fc * + + *./ - (J) Continue cancelling term until each ummation i over D(l). The coefficient of Σ A (m 1 x iι + + m t x it ) k i jut ( l) ~" f time the number of permutation on mark which are conjugate to one having cycle of length m l9, m t. Thi can be hown by a method quite imilar to that ued by Frobeniu [1]. Hence we may write Σ (18) β! Σ» = Σ (~ 1)" Σ ( ^ + + m t x h y P ) P X f where the outer ummation i over all permutation P on mark, and wfci> ' * > m a r e ί the length of the cycle of P. Now from the multinomial expanion we have Σ (m^ + + m^ Y = Σ Γ Ί V ^ m^^ 1 1 ι ).+i^kh! ih!... i t I λ S ι and the coefficient of S k i (m* + + m*)so -ι. Subtituting in (18) and uing (15), we obtain (19) (β-l)i Σ»=ΛΛ, *)&+ where the term indicated by dot do not involve S k. Thu, if f(n, k) φ 0 for k = 1,, n, then (19) can be olved recurively for Si,, S n in term of Σi> > Σn On the other hand, if f(n, k) = 0 and /(rc, j) Φ 0 f or j = 1,,Jfc - 1 then (17) expree ΣΛ a a polynomial in Si,,S fc -i which in turn are polynomial in Σi» ' * > Σt-i

7 ON THE DETERMINATION OF NUMBERS 853 COROLLARY. If f(n, k) = 0 then n divide ( - 1)! n ~\ Thu {x} will alway be determined by {σ} if i le then the greatet prime factor of n. EXAMPLE Here (18) become 6Σ* = Σ K + α? la + α? l3 )* - 3 Σ (2a? tl + α?j* + 2 Σ (3α? 4 ) fc. Expanding and collecting the coefficient of S k, we get f(n, k) = ri 2 - (2 fc + l)w + 2 3*- 1. Thi ha obviou zero at n 1, k = 1 w = 2, & = 1, 2 n 3, & = 2, 3. Alo, a we know from Theorem 3, there are zero at w = 6,& = 3,5. For all thee value of n the et [σ] doe not, in general, determine {x} uniquely. In addition we oberve that f(n, k) = 0 ha the olution n = 27, & = 5, 9 and n 486, fc = 9. We do not know whether for thee value of n the et {σ} determine {x} uniquely or not. However we do know that thee are the only cae left in doubt. THEOREM 5. // = 3 then f(n, k) = 0 ha olution only for k = 1, 2, 3, 5, 9. Proof. If /(?z, fc) = 0 then (20) n = 2 α 3 δ with α = 0 or 1. Subtituting (20) in j\n, k) = 0 we obtain (21) 2 α 3 δ + 2 τ - α 3*-*- 1 = 2 fc + 1. Let n be the maller zero of f(n, k) for a fixed &. zero i ri = 2 χ - α 3*" 6-1 and 6 < k - b - 1. Hence Then the other (22) 2* = - 1 (mod 3 δ ) and ince 2 i a primitive root of 3 &, (23) k = 3 & - T (mod 2 3 & " T ). But by (21) we have 3*-*-i < 2 fc < 3 a * /3 or A; < 3(6 + 1) o that 3*" 1 < k < 3(6 + 1) and hence 6 < 4.

8 854 J. L. SELFRIDGE AND E. G. STRAUS If 6 = 3 then k = 9 (mod 18) and k < 12 o k = 9. If 6 == 2 then 4 = 3 (mod 6) and k < 9 o k == 3. If 6 = 1 then fe == 1 (mod 2) and fc < 6 o k = 1, 3, 5. If δ = 0 then 4 < 3. EXAMPLE 2. = 4. Here (18) become (24) 24Σ*= Σ (% h + z h + x h + x h y Hence /(w, fc) = 0 become - 6 Σ (2x h + x h + x h f + 8 Σ (3α? 4l + x i ) k «Γ*2 '3 Ί'Ί + 3 Σ (2a? 4l + 2x h Y - 6 Σ (4α? 4 )*. V' λ 4 2 (25) n 3-3(2 fc " 1 + l)n 2 + (2(3* + 1) k ' τ )n *" 1 = 0. We firt note that thi ha olution n = 1, k 1 ) n 2 f fc = 1, 2; w = 3, 4 = 1, 2, 3 n = 4, & = 2, 3, 4 n = 8, fc = 3, 5, 7. For thee value of n, the et {σ} doe not generally determine {x}. When n = 12, k = 6 i a olution, and thi cae i left in doubt. THEOREM 6. If 4 then f(n f k) = 0 &a olution only for n 1, 2, 3, 4, 8, 12. Proo/. Let n = 3 α 2 δ where α = 0 or 1. Now if n > 3(2 fc " 1 + 1) then 2 3 k n > 3 fc+1 2* > *" 1 and the left ide of (25) i poitive. Hence n < 3(2 fc - χ + 1)< 2 fc+1 if k > 3 and o b < k. (For k < 3 we have lited all olution of (25)). If k i even then 2(3* + 1) == 4 (mod 8) and if fc > 4 then 8ra divide the other term unle b < 2. Similarly if & i odd then 2(3* + 1) == 8 (mod 16) and if k > 5 then b < 3. So b < 3 in all cae. Now uppoe a = 1. Then (25) become 2rc *- 1 = 0 (mod 9) or 2 δ+1 ΞΞ 2 2 *- 1 == 2 (mod 3) and 6 i even. Thu n mut be 1, 2, 3, 4, 8, or 12. It i eay to check that none of thee i a root for k > 7. The corollary to Theorem 4 how that exceptional pair (, w) are in a certain ene quite rare. Of coure it i trivial to remark that if (, n) i exceptional, then (n, w) i exceptional. Hence the remark for = 2 apply equally well to = w 2 and we obtain the exceptional pair (6, 8), (14, 16), (30, 32),. But there are other cae with n > 2 which our method leave in doubt,

9 ON THE DETERMINATION OF NUMBERS 855 THEOREM 7. We can contruct arbitrarily large value of uch that f(n, k) = 0 for ome n > 2. Proof. If n < then Σ fc = 0 but Si,,» may be precribed arbitrarily. Hence the coefficient of S k in the expanion of Σfc mut be zero if k < n. If n = then Σ* = * but S 2, -- -> S n may be precribed arbitrarily. Hence w = i a zero of /(w, A:) for k 2,, n. Thu /(, 1) = Πί:ϊ (n - i) /(rc, 2) = Πί-ifa ~ *) and/(w, 3) - (n - 2) Πf- (w - i). If we divide f(n 9 4) by it known factor then we obtain for > 2 (26) /(rc, 4) - [^2 - (6 - l)n ] Π (n - i) 4 = 4 and the equation (27) n 2 - (6 - l)n = 0 can be rewritten (2n I) 2-3(2 - I) 2 = - 2. The Pell equation ^2 3v 2 = 2 ha the general olution u + n/3~ = ± (1 + i/3~)(2 + i/3") r r = 0, ± 1,. Since u and v are odd, n and are integer. It i intereting that all poitive olution are obtained in the following imple way. When k = 4, (, n) = (2, 8) i a olution. Hence (6, 8) i a olution and putting 6 in (27) yield (6, 27). Continuing in thi way, we obtain (21, 27), (21, 98), (77, 98), (77,363),.-.. In a imilar manner we obtain for > 3 (28) j\n, 5) = O 2 - (12-5)n + I2 2 ](n - 2) Π (n - i) 4 = 5 and all integer root of the quadratic factor may be obtained with the aid of the general olution of the Pell equation u z 6v* = 75. Or we could tart with (2,16) and obtain ucceively (14,147), (133, 1444), Starting with (3, 27) yield (24, 256), (232, 2523), Concluding remark. If we let {r} = {τ lf «-, τ ns } be the et of um of not necearily ditinct element of {x}, then {x} i alway determined by {r}. A method imilar to the proof of Theorem 4 applie with the coefficient of S k alway poitive. Alternatively, if the x i are real, x x < x % < < x n, we may determine them ucceively by a imple induction procedure. Our method i applicable to the cae of weighted um σ t.. g =

10 856 J. L. SELFRIDGE AND E. G. STRAUS Σ5-i α j#i/ The reulting Diophantine equation will however be of a rather different nature. Thu, if the a ό are all ditinct then the analogue to f(n, k) = 0 i (29) (αf + a\ + + a^n 8-1 = 0. In other word the uniquene condition i independent of n and depend on the a t alone. For example if α L + a a = 0 then {σ} remain unchanged if we add the ame contant to all x. It i not a eay to ee what happen if (29) hold for ome k > 1. REFERENCES 1. G. Frobeniu, Ueber die Charaktere der ymmetrίchen Gruppe, S.-B. Preu. Akad. Wi. Berlin, (1900), L. Moer, Problem E 1248, Amer. Math. Monthly (1957), 507. UNIVERSITY OF CALIFORNIA, LOS ANGELES

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