MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems. 3. Your proofs should e net nd legile. You my nd should use the ck of pges for scrp work. 4. You re welcome to use ny non-humn sources you choose. (Note, tlking to someone on the phone, over the internet, etc. flls under getting humn help. Be sure to cite your sources s uncknowledged sources constitute plgrism nd will receive 0 points. PROBLEM POINTS SCORE 1 11 10 3 1 4 10 5 10 6 10 7 15 8 10 9 1 TOTAL 100
1.(6+5 points We proved in clss tht given integers 1,, m 1, m with gcd(m 1, m 1, then there is unique simultneous solution x modulo m 1 m to the system of equtions x 1 (mod m 1 x (mod m. ( If gcd(m 1, m d, prove tht there is simultneous solution x to the system of equtions x 1 (mod m 1 x (mod m if nd only if 1 (mod d. Show tht this solution x is unique modulo m 1 m /d. Proof: We know tht x 1 + m 1 t is solution to x 1 (mod m 1 for ny t Z. We now wnt to determine when we cn choose t Z so tht x 1 + m 1 t is lso solution to the congruence x (mod m. We hve tht x is solution to this second congruence if nd only if 1 + m 1 t (mod m, which is equivlent to m 1 t 1 (mod m. This is now liner congruence nd we know this hs solution if nd only if gcd(m 1, m ( 1, s desired. Suppose now tht x nd y re oth solutions to the system of congruences. Then x y(mod m 1 nd x y(mod m. Thus, we must hve the lest common multiple of m 1 nd m divides x y, i.e., m 1 m /d divides x y. This is precisely wht it mens for the solution to e unique modulo m 1 m /d. ( Find the smllest positive simultneous solution to the system of equtions: x 17(mod 65 x 4(mod 0. From prt ( we see tht we need to find solution to 65t 4 17(mod 0, i.e., solution to 5t 5(mod 0. This is ovious though with t 1. Thus, our simultneous solution is x 17 + 65 8.
3. (10 points Let D e positive integer nd suppose there exists prime p dividing D so tht p 3(mod 4. Prove tht x Dy 1 hs no integer solutions. Proof: Suppose x 0, y 0 is n integer solution to the eqution, i.e., x 0, y 0 Z nd Looking t this eqution modulo p we otin x 0 dy 0 1. x 0 1(mod p. This implies tht 1 is qudrtic residue modulo p. Since p 3(mod 4 this cnnot e the cse. Thus, there re no integer solutions to this eqution. 3. (6+6 points ( Evlute the continued frction [ : 1, ]. Set x [1, ]. Then we hve x [1,, 1, ]. From this we hve This leds to the qudrtic eqution x 1 + 1 + 1 x 1 + x x + 1 3x + 1 x + 1. x x 1 0. We cn use the qudrtic eqution to solve for x, noting tht x > 0 to discrd one of the solutions: So we hve x 1 + 3. [ : 1, ] + 1 x + 1 + 3 + 3 1 3 1 + 3.
4 ( Express 71 s finite simple continued frction. Do this y hnd! We use the Eucliden lgorithm to ccomplish this: 71 (1 + 16 16(3 + 7 16 7( + 7 (3 + 1 1( + 0. We now convert these equtions into the form we need: 71 1 + 16 16 3 + 7 16 16 7 + 7 7 3 + 1 + 0. Comining these equtions we otin the following continued frction: 71 1 + 16 1 + 1 16 1 + 1 3 + 7 16 1 + 1 3 + 1 16 7 1 1 + 1 + Thus, the continued frction expnsion of 71 3 + 1 + 7 1 3 + 1 + 1 3 + 1 is [1; 3,, 3, ]..
5 4. (10 points Is 137 congruent numer? Be sure to justify your nswer. If it is congruent numer, find tringle with rtionl sides nd re 137. Using SAGE one finds the rtionl point ( 3136 5, 7711 15. This is clerly not in the set E137 (Q tors, so 137 is congruent numer. Previous homework llows one to turn this rtionl point into the tringle with side lengths: 1377 80, 7670 1377, 156511 3860, which one cn esily check hs re 137. 5. (10 points Prove tht 3 is primitive root of ll integers of the form 7 k nd 7 k for k 1. Proof: For this one needs to go ck to how we proved primitive roots existed for prime powers. In fct, with the proper set-up one could just quote these results. It is elementry to check tht 3 is primitive root modulo 7. We cn lso check tht 3 6 1(mod 49. Thus, 3 stisfies the hypotheses of Lemm on pge 160 of the textook (we lso covered this in clss, ut we didn t numer it there. Thus, we hve 3 7k 6 1(mod 7 k for ll k. Now the proof of Theorem 8.9 of the text shows tht 3 is primitive root for 7 k with k 1 nd the Corollry to Theorem 8.9 shows 3 is primitive root of 7 k for ll k 1. 6. (10 points Find 6 different positive vlues of n so tht n+1 nd n +1 re oth perfect squres. Are there infinitely mny different vlues for n so tht n + 1 nd n + 1 re oth perfect squres? Prove your nswer is true. (You my use ny theorems proven in clss to prove your result. Proof: Suppose n + 1 nd n + 1 re oth perfect squres, i.e., there exists x, y Z so tht n + 1 x nd n + 1 y, i.e., n + y. Sutrcting these two equtions gives tht n + 1 nd n + 1 re perfect squres if nd only if there is n integer solution to the eqution x y 1. We know from clss tht if we let p k q k e the convergents of the continued frction expnsion of, then p k q k ( 1k+1 k+1 where k+1 is defined s in clss. We sw tht this eqution hs solutions if nd only if the period of the continued frction expnsion of is odd, which in this cse it is with period n 1. We found the solutions to e given y p n(t+1 1, q n(t+1 1 with t 1, which in our cse re given y p t, q t. This shows in fct tht we hve infinitely mny x, y Z stisfying this eqution nd thus infinitely mny such n. We now just need to find 6 of them. The first 6 solutions of the eqution re: (x, y (7, 5, (41, 9, (39, 169, (1393, 985, (8119, 5741, (4731, 33461. The corresponding vlues of n re: 48, 1680, 5710, 1940448, 65918160, 3977040.
6 7. (5 points ech Let nd > 1 e reltively prime integers with odd. Write p 1 p r with the p i odd, define the Jcoi symol ( y p 1 ( Prove tht if is qudrtic residue of then 1. Is the converse true? Prove it or give counterexmple. Proof: Let p 1 p r with p i odd primes. If is qudrtic residue modulo then there exists n integer x so tht x (mod. Thus, (x. Since p i, we hve x (mod p i for ech i. Nmely, we hve n ( n 1 1. i1 p i The converse is flse. Consider 3 nd 35. Then is not qudrtic residue modulo 35 (check y computtion ut ( ( ( 3 3 3 ( 1( 1 1. 35 5 7 ( Prove tht ( ( nd ( (. p r i1. Proof: Let p 1 p r. We hve ( n ( i1 p i n ( ( i1 p i p i where we hve used the nlogous property for the qudrtic residue symol. Let p r+1 p s with p j (r + 1 j s primes. We hve s climed. s ( i1 p i r ( s i1 p i jr+1 ( p j
7 (c Prove tht if nd re reltively prime positive odd integers ech greter then 1, then ( ( 1 1 1. It my e helpful to note tht if u, v re odd integers, then (u 1/+(v 1/ (uv 1/(mod. Proof: Let p 1 p r nd q 1 q s. We hve s ( i1 s q i ( pj r q i1 j1 i s r ( qi i1 j1 p j ( 1 pj 1 ( s r ( 1 i1 j1 ( s i1 ( 1P ( s qi 1 i1 ( 1P ( ( 1 ( 1 1 pj 1 P r j1 pj 1 qi 1 qi 1 qi 1 ( 1P r j1 pj 1 (qudrtic reciprocity where the lst equlity follows from the hint. Now we use tht ( ( ±1 to move to the ( other side of the eqution ( 1. 8. (10 points Prove tht φ(p! (p 1φ((p 1! for p prime. Proof: We cn write p! p(p 1!. Since p is prime, gcd(p, (p 1! 1 nd so φ(p(p 1! φ(pφ((p 1! (p 1φ((p 1!, s climed.
8 Choose either version of prolem 9 to do. You my receive extr credit for doing the other one though! 9. (1 points Consider the eqution + p(c + d. Show tht if p 3(mod 4 tht this eqution hs no solutions. (You my wnt to use descent here! Proof: Let u 1, v 1, x 1, y 1 Z >0 e solution to the eqution so tht u 1 + v 1 p(x 1 + y 1. (1 This implies tht u 1 + v 1 0(mod p, ( i.e., u 1 v 1 (mod p. Suppose p v 1. Then we hve tht v 1. However, we know tht 1p 1 nd so we must hve 1. This is impossile ( v 1 p since p 3(mod 4. Thus, we hve p v 1 nd hence p u 1 s well. Set u u 1 /p nd v v 1 /p. Then we hve (pu + (pv p(x 1 + y 1 i.e., x 1 + y 1 p(u + v. Note tht u < u 1. We cn now repet the sme rgument to otin tht p x 1 nd p y 1. Setting x x 1 /p nd y y 1 /p nd pplying the sme rgument we otin u + v p(x + y. However, this gives nother solution (u, v, x, y with strictly smller positive vlues then our originl solution. Applying descent we see tht we hve strictly decresing sequence of positive integers, contrdiction. Thus there cn e no solutions. ( 1 p
9 9. (6+6 points Let m e positive squre-free integer. ( Show tht the elements + m with, Z re ll lgeric integers in Q( m. Proof: Set α + m with, Z. Oserve tht α + α Z nd αα m Z. Thus, the polynomil f(x x (α + αx + αα is monic polynomil with integer coefficients. Plugging in α we see tht f(α α (α + αα + αα 0. Thus, we see tht α is necessrily n lgeric integer in Q( m. ( Prove there re infinitely mny units in Q( m. Proof: Recll tht α + m Q( m is unit if nd only if N(α αα ±1 nd α is n lgeric integer. The sttement out the norm is equivlent to the sttement tht m ±1. Consider the eqution x my 1. This is Pell s eqution nd we know there re infinitely mny solutions to this eqution for m positive nd squre-free. For ny solution x n, y n of the eqution x my 1 we hve tht the element x n +y n m is then unit in Q( m y prt ( long with the fct tht it hs norm 1. Since there re infinitely mny integer solutions to Pell s eqution, there re infinitely mny units in Q( m.