5. Assume $p \nmid a$. Show that $x^n \equiv a \pmod{p}$ is solvable if and only if \\ $a^{\frac{p-1}{(n, p-1)}} \equiv 1 \pmod{p}$ \\ and there are exactly $(n, p - 1)$ solutions $\pmod{p}$. (Hint: Follow the proof of Lemma 2.3.)
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Then $n = dn_1$ and $p-1 = dd_1$ for some integers $n_1$ and $d_1$. Show more…
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Let a, n ∈ Z^+ with d = gcd(a, n). Show the equation ax (mod n) = 1 has a solution iff d = 1.
Adi S.
(1) Show that if a and n are relatively prime integers with n > 1, then n is prime if and only if (x - a)^n and x^n - a are congruent modulo n as polynomials. (Hint: Let q be a prime divisor of n and suppose q^k is the largest power of q which divides n. Consider the coefficient of x^q in the binomial expansion of (x - a)^n and show that it is not divisible by n.)
Sri K.
13. This theoretical exercise is intended to provide some insights into the number of solutions of the congruence x^2 ≡ 1 mod n. We can suppose n > 2, since for n = 2 the only solution is x ≡ 1 mod 2. For n > 2, there are two obvious solutions: x ≡ ±1 mod n. Depending on n there could be more solutions. For instance, with n = 8 we also have the solutions x ≡ ±3 mod 8. We might notice another tidbit here. If a solves such congruences, so does −a simply because (-a)^2 = a^2 ≡ 1 mod n. Thus we might predict that the number of solutions to these congruences is even. (a) If p is an odd prime and x^2 ≡ 1 mod p, prove that x ≡ ±1 mod p. So, for odd primes the two obvious solutions are the only solutions. (b) Suppose m, n are coprime moduli, and that the number of solutions to the individual congruences x^2 ≡ 1 mod m, x^2 ≡ 1 mod n are r, s, respectively. Prove that the number of solutions to the congruence x^2 ≡ 1 mod mn is rs. Hint. There are no tricks, but a well-written proof requires clear thinking. To get started let a1, a2, . . . , ar be the distinct solutions to the congruence x^2 ≡ 1 mod m, and let b1, b2, . . . , bs be the distinct solutions to the congruence x^2 ≡ 1 mod n. Then use Chinese Remainder Theorem (CRT) to obtain solutions to the congruence x^2 ≡ 1 mod mn. Show that the number of solutions so obtained is rs, and that every solution arises in this fashion. Proposition 4.1 also plays a big role. (c) How many solutions does the congruence x^2 ≡ 1 mod 56 have? Proposition 4.1. If a, b are integers and m, n are coprime moduli, the congruence a ≡ b mod mn is true if and only if the pair of congruences a ≡ b mod m and a ≡ b mod n are true. Thus any congruence problem involving a modulus that is the product of co-prime integers can be reduced to a pair of congruence problems involving the factors that made up the product. We will be using this fact persistently. Theorem 4.2 (Chinese Remainder). If m, n are coprime moduli, and a, b are any integers, then the congruences x ≡ a mod m and x ≡ b mod n have a common solution x. Furthermore, any two solutions x, y to this pair of congruences must be such that x ≡ y mod mn. Proof. Since m, n are coprime, the Diophantine equation mt − ns = b − a has a solution t, s. With such t, s we obtain mt + a = ns + b. Now, let x be this common number. Clearly x ≡ a mod m and x ≡ b mod n, which makes x a solution to the simultaneous congruences. If y is another solution to the simultaneous congruences, then x ≡ y mod m and x ≡ y mod n. According to Proposition 4.1, we conclude x ≡ y mod mn.
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