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MT 430 Intro to Number Theory PROBLEM SET 3 Due Thursday 2/14 Remark. Here, as in class, whenever working modulo m, we use notation x to denote a multiplicative inverse of x modulo m. Whenever (x, m) = 1, x is the unique residue modulo m such that xx ≡ 1 (mod m). Problem 1. Let p be an odd prime number. Show that the congruence x2 ≡ 1 (mod pn ) has only the two solutions x ≡ 1 (mod pn ) and x ≡ −1 (mod pn ). Solution: The statement x2 ≡ 1 (mod pn ) is equivalent to pn | (x2 − 1) = (x − 1)(x + 1). Since (x + 1) − (x − 1) = 2, the numbers (x + 1) and (x − 1) cannot be divisible by p simultaneously. If (x − 1) is not divisible by p, then since p is a prime and pn divides the product (x − 1)(x + 1), we must have pn | x + 1. Similarly, if (x + 1) is not divisible by p, we must have pn | x − 1. Problem 2. Find all x such that x ≡ 1 (mod 4) x ≡ 2 (mod 3) x ≡ 5 (mod 7) We apply the algorithmic version of Chinese Remainder Theorem from class. Take b1 = 21 = 1 = 1 (mod 4), b2 = 28 = 1 = 1 (mod 3), and b3 = 12 = 5 = 3 (mod 7). Then x = 1 × b1 × 21 + 2 × b2 × 28 + 5 × b3 × 12 = 21 + 56 + 180 ≡ 5 (mod 84) is the unique solution of the system modulo 84 = 4 × 3 × 7. Problem 3. Solve the congruence x3 + 2x − 3 ≡ 0 (mod 45). Solution: Note that 45 = 5 × 32 . First, we solve separately the congruence x3 + 2x − 3 ≡ 0 modulo 5 and modulo 32 . The solutions of x3 + 2x − 3 ≡ 0 (mod 5) are x ≡ 1, 3 (mod 5). To find solutions modulo 32 , we first observe that modulo 3, x3 +2x−3 ≡ x+2x− 3 ≡ 3x − 3 ≡ 0 (mod 3) by Fermat’s little theorem. Hence all x ≡ 0, 1, 2 (mod 3) are all solutions modulo 3. We then compute that (x3 + 2x − 3)0 = 3x2 + 2 ≡ 2 (mod 3). Hence by Hensel’s lemma, every solution modulo 3 lifts to a unique solution modulo 9. Since 2 = 2 (mod 3), we see that • x ≡ 0 (mod 3) lifts to x ≡ 0 − 2f (0) = 0 − 2(−3) = 6 (mod 9) • x ≡ 1 (mod 3) lifts to x ≡ 1 − 2f (1) = 1 − 0 = 1 (mod 9) • x ≡ 2 (mod 3) lifts to x ≡ 2 − 2f (2) = 2 − 18 = 2 (mod 9) We summarize that solutions of x3 + 2x − 3 ≡ 0 (mod 9) are x ≡ 1, 2, 6 (mod 9). Combining these with the solution x ≡ 1, 3 (mod 5) of x3 + 2x − 3 ≡ 0 (mod 5) and applying Chinese Remainder Theorem, we obtain 6 solutions modulo 45: x ≡ 1, 28, 11, 38, 6, 33 1 (mod 45). 2 PROBLEM SET 3 Problem 4. Solve x5 + x4 + 1 ≡ 0 (mod 34 ). Hint: Hensel’s lemma is inapplicable here because f 0 (1) = 3 ≡ 0 (mod 3). You can still use the unique solution modulo 3 to look for solutions modulo 9. Solution: The only solution of x5 + x4 + 1 ≡ 0 (mod 3) is x = 1. Hence the only candidates for solutions of x5 + x4 + 1 ≡ 0 (mod 32 ) are x = 1, 1 + 3 = 4, and x = 1 + 6 = 7. We check that 15 + 14 + 1 ≡ 3 5 (mod 9) 4 4 + 4 + 1 ≡ 3 (mod 9) 75 + 74 + 1 ≡ 3 (mod 9) 5 4 Therefore, the x + x + 1 ≡ 0 has no solutions modulo 9 and thus no solutions modulo 34 . Problem 5 (Edited on 2/12). Let p be an odd prime and suppose that a is an integer such that x2 ≡ a (mod p) has a solution and a 6≡ 0 (mod p). Show that x2 ≡ a (mod pn ) has a solution for all integers n ≥ 1. Solution: The proof is by induction. The base case of n = 1 is given to us. Suppose b is a solution of x2 − a ≡ 0 (mod pn ). Since (x2 − a)0 = 2x and 2b 6≡ 0 (mod p), Hensel’s lemma says that b lifts to a solution of x2 − a ≡ 0 (mod pn+1 ). Remark. Euler’s criterion says that for a 6≡ 0 (mod p) the equation x2 ≡ a (mod p) has a solution if and only if x p−1 2 ≡1 (mod p). Assuming Euler’s criterion, the old statement of Problem 5 is equivalent to the new statement.