PROBLEM SET 5 SOLUTIONS. Solution. We prove that the given congruence equation has no solutions. Suppose for contradiction that. (x 2) 2 1 (mod 7).

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1 PROBLEM SET 5 SOLUTIONS 1 Fid every iteger solutio to x 17x 5 0 mod 45 Solutio We rove that the give cogruece equatio has o solutios Suose for cotradictio that the equatio x 17x 5 0 mod 45 has a solutio Sice 45 is divisible by 7, there must exists a solutio to the cogruece equatio x 17x 5 0 mod 7 We may simlify this equatio to x 4x 5 0 mod 7, which ca be also writte as This imlies that 1 However, we have 7 desired cotradictio, comletig the roof x 1 mod 7 1 sice 7 3 mod 4 Hece we obtai a 7 Let be a odd rime, ad a, N, with a Cosider the equatio x a mod a Show that if, 1 1, a solutio always exists Solutio Let g be a rimitive root modulo Note that x 0 mod is ot a solutio sice a 0 mod Hece we may write x g k mod ad a g l mod for some itegers k ad l The we have x g k g k mod, so we ca rewrite the give equatio as g k g l mod Sice the order of g is 1, this is equivalet to k l mod 1

2 We ca solve this cogruece equatio for k, amely k 1 l mod 1 where 1 is the iverse of modulo 1 Note that 1 exists because, 1 1 Hece we fid a solutio x g k mod to the give equatio x a mod b Assume that 1 mod Prove there exists a solutio if ad oly if a 1 mod Solutio Suose first that we have a solutio to the equatio x a mod The we have x 0 mod as already oted i art a Hece we immediately see that a x x 1 mod where the last cogruece follows from Fermat s little theorem For the coverse, suose ow that a 1 mod As i art a, take a rimitive root g modulo ad write a g l mod for some iteger l The we have l g g l a 1 mod Sice the order of g modulo is 1, we deduce that l 1 is divisible by 1, or equivaletly that l is divisible by If we write l t for some iteger t, we have g t g t g l a mod Hece the equatio x a mod has a solutio, amely x g t mod 3 Let a, m > 1 be odd itegers with a, m 1 Suose that the rime factorizatio of a is a α1 1 α αr r q β1 1 qβ qβs s where each i 1 mod 4 ad each q j 3 mod 4 Put b solutio for all rimes m s β j Assume x mod a has a j1 a Prove that if 1 mod 4 for all rimes m, the x a mod m has a solutio Solutio We begi with the followig stadard lemma: Lemma The cogruece equatio x a mod m has a solutio if ad oly if rimes m 1 for all

3 Proof Let us first rove the oly if art If the equatio x a mod m has a solutio, for all rimes m the equatio x a mod also has a solutio, meaig that 0 or 1 However, a 0 mod for all rimes m sice a ad m are relatively rime Hece we deduce that 1 for all rimes m For the if art, we assume that 1 for all rimes m Our goal is to show that the equatio x a mod m has a solutio For each m, let e be the exoet of i the rime factorizatio of m By the Chiese Remaider Theorem, it suffices to show that the equatio x a mod e has a solutio for all rimes m Take fx x a Note that the equatio fx 0 mod has a solutio sice 1 We have f x x 0 mod if ad oly if x 0 mod However, x 0 mod is ot a solutio to fx 0 mod sice a 0 mod as see i the first aragrah Hece Hesel s lemma tells us that the equatio fx 0 mod e has a solutio We ow retur to the roblem Fix a rime dividig m By Lemma, it suffices to show that 1 By multilicity of Legedre symbols, we have a α1 1 α αr r 1 α1 α q β1 1 qβ qβs s r αr q1 β1 β q βs qs Let us ow comute each Legedre symbol i the above equatio Sice the equatio x mod a has a solutio, the equatios x mod i ad x mod q j have a solutio for all i ad j We also have, i, q j 1 for all i ad j sice a ad m are relatively rime Hece we fid that 1 for all i 1,,, r ad j 1,,, s i q j Now we aly ad the quadratic recirocity law to obtai i i i q j 1 i 1 for all i 1,,, r, 1 q j 1 for all j 1,,, s 1

4 For each i 1,,, r the umber i 1 is eve sice i 1 mod 4 O the other had, for each j 1,,, s the umber q j 1 is odd sice q j 3 mod 4 Hece we obtai i 1 for all i 1,,, r, 1 for all j 1,,, s Note that 1 is a eve umber sice 1 mod 4 Hece we have i 1 for all i 1,,, r ad j 1,,, s By 1 we deduce that 1, thereby comletig the roof 3 b Prove that if 3 mod 4 for some rime m, the x a mod m has a solutio if ad oly if b is eve Solutio By Lemma, the equatio x a mod m has a solutio if ad oly if 1 for all a rimes m O the other had, we roved i art a that 1 whe 1 mod 4 Hece we deduce that the equatio x a mod m has a solutio if ad oly if 1 for all rimes m with 3 mod 4 Let us ow fix a rime m such that 3 mod 4 Note that 1 ad 3 remai valid However, 1 is ow a odd umber sice 3 mod 4 Hece 3 yields i 1 for all i 1,,, r, 1 for all j 1,,, s We substitute these umbers to 1 ad fid that 1 α1 1 α 1 αr 1 β1 1 β 1 βs 1 β1β βs 1 b Hece we coclude that 1 if ad oly if b is eve We thus comlete the roof by the discussio i the first aragrah

5 4 Let a be a o-zero iteger For a rime, let deote the umber of mod iteger solutios x, y of the equatio y x 3 ax mod a Show that for a odd rime, r 3 r1 Solutio For a fixed umber r {1,,, }, the air r, y is a solutio if ad oly if y r 3 mod r 3 Note that the umber of y mod satisfyig this equatio is 1 Hece we have r1 [ 1 b Prove that if 3 mod 4, r 3 ] r 3 r 3 Solutio Note that the value of deeds oly o r mod Hece we may write r 3 r1 r 3 r1 r1 1 r 3 r1 r r 3 [ r 3 r1 r1 r 3 r 3 a r r 3 ] r1 r 1 r 3 3 a 3 a 3 a 3 a Note that the last term vaishes sice 3 a 0 mod Hece it suffices to rove the followig: r 3 r 3 I fact, this follows from the fact that r for each r 1,,, 1 1 if 3 mod 4; more recisely, we have r 3 r 3 for each r 1,,, 1