# MTH 505: Number Theory Spring 2017

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1 MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of \$ nd \$ s two denomintions of coins nd \$c s some vlue tht we wnt to py. The eqution hs solution px, yq P N 2 if nd only if we cn mke chnge for \$c, nd in this cse we sy tht c is p, q-representle. More generlly, we will consider the set of p, q-representtions of c: R,,c : tpx, yq P N 2 : x ` y cu. The prolem is trivil when 0 so we will lwys ssume tht 0, i.e., tht nd re oth nonzero. 2.. Consider nturl numers,, c P N with d gcdp, q, where d nd d. () If d c prove tht R,,c H. () If d c with c dc prove tht R,,c R,,c. [Unlike the cse of Diophntine equtions, it is possile tht oth of these sets could e empty.] Proof. (): Let d c nd ssume for contrdiction tht R,,c is not empty, i.e., ssume tht there exists pir of nturl numers px, yq P N 2 such tht x ` y c. But then we hve which contrdicts the fct tht d c. c x ` y pd qx ` pd q dp x ` yq, (): Now suppose tht d c so tht c dc for some c P Z. Since c nd d re oth positive we must hve c P N. To show tht R,,c Ď R,,c consider ny px, yq P N 2, so tht x` y c. Then we hve x ` y c dp x ` yq dpc q pd qx ` pd qy pdc q x ` y c, which sys tht px, yq P R,,c s desired. Conversely, consider ny px, yq P R,,c, so tht x ` y c. Then we hve x ` y c pd qx ` pd qy pdc q dp x ` yq dpc q x ` y c, which sys tht px, yq P R,,c. (The lst step used multiplictive cncelltion.)

2 The previous result llows us to restrict our ttention to coprime nd Let,, c P N with 0 nd gcdp, q. If R,,c H (i.e., if c is p, q-representle) prove tht there exists unique representtion pu, vq P R,,c with the property 0 ď u ă. [Hint: For existence, let px, yq P R,,c e n ritrry solution. Since 0 there exists quotient nd reminder of x mod. For uniqueness, use the coprimlity of nd to pply Euclid s Lemm.] Proof. If R,,c H then there exists some pir px, yq P N 2 such tht x ` y c. Since 0 there exists pir of integers q, r P Z such tht " x q ` r Then sustituting x q ` r gives 0 ď r ă x x ` y c pq ` rq ` y c r ` pq ` yq c. It only remins to check tht pu, vq : pr, q ` yq P N 2 nd we lredy know tht r P N. Since r ă x we lso hve q px rq ą 0, which since ą 0 implies tht q ą 0. But then since y P N we hve q ` y P N s desired. This proves existence. For uniqueness, ssume tht we hve pu, v q nd pu 2, v 2 q in R,,c with 0 ď u ă nd 0 ď u 2 ă. Then since u ` v c u 2 ` v 2 we see tht u ` v u 2 ` v 2 pu u 2 q pv 2 v q, which implies tht divides pu u 2 q. But then since gcdp, q, Euclid s Lemm sys tht pu u 2 q. If pu u 2 q 0 then we re done. Otherwise, suppose without loss of generlity tht u u 2 ą 0. Then the fct tht pu u 2 q implies tht ď u u 2 ď u which contrdicts the fct tht u ă. This contrdiction shows tht pu u 2 q 0 nd then the eqution pv 2 v q pu u 2 q 0 0 together with the fct 0 implies tht pv 2 v q 0 s desired Let, P N e coprime with 0. If c p q prove tht R,,c H. Tht is, prove tht the numer p q is not p, q-representle. [Hint: Let c p q nd ssume for contrdiction there exists representtion px, yq P R,,c. Show tht the cses x ă nd x ě oth led to the contrdiction y ă 0. You cn use 2.2 for the cse x ă.] Proof. Assume for contrdiction tht we hve representtion x ` y p q with px, yq P N 2. From 2.2 this implies tht there exists representtion u ` v p q

3 with pu, vq P N 2 nd 0 ď u ă. Now oserve tht u ` v u ` v pu ` q p vq. The lst eqution sys tht divides pu ` q nd then since nd re coprime we otin pu ` q from Euclid s Lemm. Since u ` ą 0 this implies tht ď u ` [this rgument is in the notes] ut we lredy know tht u ă (i.e., u ` ď ) so we conclude tht u `. Finlly, we sustitute u to otin which contrdicts the fct tht v P N. u ` v p q ` v ` v v v, [Sorry I didn t follow my own hint very closely.] 2.4. Let, P N e coprime with 0. In this exercise you will prove y induction tht every numer c ą p q is p, q-representle. () Prove the result when or. () From now on we will ssume tht ě 2 nd ě 2. In this cse prove tht the numer p ` q is p, q-representle. [Hint: From the Eucliden Algorithm nd 2.2 there exist x, y P Z with x ` y nd 0 ď x ă. Prove tht px q P N nd py ` q P N, nd hence is vlid representtion.] px q ` py ` q p ` q (c) Let n ě p ` q nd ssume for induction tht n is p, q-representle. In this cse prove tht n ` is lso p, q-representle. [Hint: Let x, y e s in prt (). By the induction hypothesis nd 2.2 there exist x, y P N with x ` y n nd 0 ď x ă. Note tht px ` x q ` py ` y q pn ` q. If y ` y ě 0 then you re done. Otherwise, show tht is vlid representtion.] px ` x q ` py ` y ` q pn ` q Proof. (): Since the prolem is symmetric in nd we will ssume without loss of generlity tht. Now we wnt to show tht every numer c ą p q, i.e., every numer c ě 0 is p, q-representle. But this is certinly true ecuse p0q ` p0q 0 is vlid representtion of c 0 nd pq ` pc q c is vlid representtion of c ą 0. This solves the prolem when or so from now on we will ssume tht ě 2 nd ě 2. (): Bse Cse. Since gcdp, q the Eucliden Algorithm gives integers x, y P Z such tht x ` y nd from 2.2 we cn ssume tht 0 ď x ă. [Actully this is it esier

4 thn 2.2 ecuse we don t require y ě 0.] If x 0 then we hve y x ` y which implies tht, contrdicting the fct tht ě 2. Thus we must hve x ě, i.e., x P N. To complete the proof, ssume for contrdiction tht py ` q ă 0, i.e., y ` ď 0. This implies tht y ď nd hence y ď. Finlly, since px q ă 0 we otin the desired contrdiction: x ` y ď x px q ă 0. We conclude tht px q nd py ` q re nturl numers, so px q ` py ` q px ` y q ` ` is vlid p, q-representiton of p ` q. (c): Induction Step. Let n ě p ` q nd ssume for induction tht there exist nturl numers px, yq P N 2 such tht x ` y n. In this cse we wnt to show tht n ` is lso p, q-representle. To do this, recll from prt () tht we hve integers x, y P Z with the following properties: x ` y, ď x ď, y ` ě. Now dd the equtions x ` y n nd x ` y to otin px ` x q ` py ` y q n `, where x ` x ě 0. If we lso hve y ` y ě 0 then we re done, so ssume tht y ` y ă 0. Since y ` ě nd y ě 0 we hve py ` y ` q ě. It only remins to check tht px ` x q ě 0. To see this we use the ssumptions pn ` q ě p ` 2q nd py ` y ` q ď 0 to otin n ` px ` x q ` py ` y q ą ` 2 px ` x q ě py ` y q ` 2 ą py ` y ` q ` 2 ě p0q ` 2 ą p q ą 0. By cncelling ą 0 from oth sides of px ` x q ą p q we otin px ` x q ą p q nd hence px ` x q ě 0 s desired. It follows tht px ` x q ` py ` y ` q px ` yq ` px ` y q ` p ` q n ` ` 0 is vlid p, q-representtion of n `. [Tht ws tricky.] Let, P N e coprime with 0. So fr you hve proved tht R,,p q 0 nd R,,c ě for ll c ą p q. The next prolem gives rough lower ound for the totl numer of p, q-representtions.

5 2.5. Let, P N e coprime with 0. Prove tht Y c R,,c ě mxtn P N : n ď c{pqu. ] [Hint: We know from clss tht the integer solutions of x ` y c hve the form px, yq pcx k, cy ` P Z, where x, y P Z re some specific integers stisfying x `y. Now prove tht the nturl numer solutions correspond to vlues of k P Z such tht cy ď k ď cx. Counting these integers is delicte ut you should e le to give rough ound.] Proof. Consider,, c P N with gcdp, q. From 2.2 there exist integers x, y P Z such tht x ` y nd 0 ď x ă. We know from clss tht the complete integer solution to the eqution x ` y c is given y px, yq pcx k, cy ` P Z, nd our jo is to discover which of these solutions re non-negtive. Tht is, we need to find ll integers k P Z such tht the following two inequlities hold: cx k ě 0 cy ` k ě 0. These two inequlities cn e written in terms of frctions to otin cy ď k ď cx. Ech such vlue of k P Z corresponds to non-negtive solution of x`y c, so we conclude tht R,,c is equl to the numer of integers in the closed rel numer intervl r cy {, cx {s. The exct count is tricky, ut the floor of the length of the intervl provides lower ound: Z ^ cx R,,c ě cy Z cx ` cy ^ Z cpx ` y ^ q Y c ]. Unfortuntely this rough ound gives us no informtion when c ă, i.e., when tc{pqu 0. With it more work one could compute the exct formul: for ny x ` y we hve ( ) R,,c c " " cy cx `, where txu : x txu is the frctionl prt of the rtionl numer x P Q. This formul is due to Tieriu Popoviciu in 953.

6 2.6. Let, P N e coprime with 0. Given n integer 0 ă c ă such tht c nd c, use Popoviciu s formul ( ) to show tht R,,c ` R,,p cq. [Hint: Use the fct tht t xu txu when x R Z.] Proof. Consider,, c P N with gcdp, q, 0 ă c ă, nd where nd do not divide c. Then for ny integers x ` y Popoviciu s formul gives R,,p cq c " " p cqy p cqx ` 2 c "y cy "x cx. But now oserve tht for ll integers n P Z nd non-integer rtionls x P Q we hve tn xu t xu txu. Thus the ove formul ecomes R,,p cq 2 c "y cy "x cx 2 c ˆ " ˆ " cy cx ˆ " " c cy cx ` R,,c. In conclusion, one cn show from 2.6 tht there exist exctly p qp q 2 nturl numers tht re not p, q-representle. This fct ws first proved y Jmes Joseph Sylvester in 884. Proof. I didn t sk you to show this, ut here s the proof. Let gcdp, q. Then we know tht every integer c ě is p, q-representle. [In fct we know tht every integer c ą p q is representle, ut we don t need this right now.] Of the ` elements of the set tc P Z : 0 ď c ď u we know tht elements re multiples of, nd elements re multiples of. Furthermore, since gcdp, q we know tht the only elements tht re multiples of oth nd re 0 nd. We conclude tht there re exctly p ` q p ` 2q p ` q p qp q elements of the set tht re not multiple of or. The result of Prolem 2.6 sys tht exctly hlf of these numers re p, q-representle. Epilogue: The proofs ove re lgeric, ut there is lso eutiful geometric wy to think out the Froenius coin prolem. Consider, P N with 0 nd gcdp, q. Lel ech point px, yq P Z 2 of the integer lttice y the numer x ` y. Note tht points on the sme line of slope { receve the sme lel. The prolem is to count the integer points on the line x ` y c tht lie in the first qudrnt. For exmple, here is the lelling corresponding to the coprime pir p, q p5, 8q:

7 I hve drwn the lines 5x ` 8y nd 5x ` 8y 0. It ws reltively esy to show tht every lel c ě 40 occurs in the first qudrnt, ut the numers elow 40 re more tricky. I hve outlined the numers elow 40 re re not multiples of 5 or 8 ut re still p5, 8q-representle. We oserve tht there re p5 qp8 q{2 4 such numers, s expected. I hve lso outlined the numers in the fourth qudrnt tht re not p5, 8q-representle. Oserve tht these two shpes re congruent up to 80 rottion, nd in fct this is the trnsformtion c ÞÑ p cq. Oserve further tht the two shpes fit together perfectly to mke n p q ˆ p q rectngle. This is the geometric explntion for Sylvester s formul p qp q. 2

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