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, qrepresentle. More generlly, we will consider the set of p, qrepresenttions 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, qrepresentle) 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, qrepresentle. [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, qrepresentle. () 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, qrepresentle. [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, qrepresentle. In this cse prove tht n ` is lso p, qrepresentle. [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, qrepresentle. 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, qrepresentiton 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, qrepresentle. 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, qrepresenttion 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, qrepresenttions.
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 nonnegtive. 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 nonnegtive 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 noninteger 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, qrepresentle. 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, qrepresentle. [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, qrepresentle. 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, 8qrepresentle. 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, 8qrepresentle. 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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