Fermat's Little Theorem says that for all natural numbers p and integers x,if p is prime then x² = x mod p.(i) Show that the statement is wrong, if we drop the assumption that p is prime.[Remember: All you need to do is to give a counterexample.](ii) Show, by giving an example, that if p is not prime, then x² = x modp can stillbe true for some integers x with x ‡ −1, 0, 1.

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Answered: Fermat's Little Theorem says that for…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Prove the following statements (using either direct or indirect proof method) state method used (a) For every integer n, n3 + n is even. (Hint: An integer can be either odd or even. Use both cases) (b) The product of two integers, one of the form 3m1 + 1 and the other of the form 3m2 + 2, where m1 and m2 are integers, is of the form 3m3 + 2 for some integer m3.
. (a) If g is a primitive root modulo 37, which of the numbers g², g³,..., g8 are primitive roots modulo 37?
Let p and q be prime numbers.(1) If p is not equal to q, how many generators does Zpq have?(2) If p = q, how many generators does Zp2 have
Suppose that P(n), Q(n) and R(n) are statements about the integer n. Let S(n) be the statement: “If P(n) is true then Q(n) and R(n) are both true." (a) The contrapositive of S(n) is the statement Olf Q (n) and R(n) are both true then P(n) is false. Olf Q (n) and R(n) are both true then P(n) is true. Olf at least one of Q(n) and R(n) is false then P(n) is false. Olf Q (n) and R(n) are both false then P(n) is false. Olf at least one of Q(n) and R(n) is false then P(n) is true. Olf P(n) is false then at least one of Q(n) and R(n) is false. Olf Q (n) and R(n) are both false then P(n) is true. ONone of the other answers. (b) The statement: "If Q (n) and R(n) are both true then P(n) is true." is OThe converse of S(n). OThe contrapositive of S(n). OThe converse of the contrapositive of S(n). ONone of the other answers.
1)For each of the following, find an integer value x which satisfies the equation, or explain why there is no solution to the equation. (a) 2x ≡ 3(mod 25) (b) 6x ≡ 4(mod 15)
Put the steps of a proof for the following claim in the proper order: 4 + 8 + 12 + + 4n = 2n(n + 1) + 4k + 4(k + 1) = 2k (k + 1) +4(k + 1) 2 (k + 1) (k +2) • 4+8+ 12 + .. + 4k = 2k (k + 1). Assume 4 + 8 + 12 + v Let n = 1. Clearly, 4 = 2(1)(1+ 1). v WTS: 4 +8+ 12 + + 4k + 4(k + 1) = 2 (k + 1) ((k + 1) + 1)
Fermat's Little Theorem says that for all natural numbers p and integers x, if p is prime then xº = x mod p. (i) Show that the statement is wrong, if we drop the assumption that p is prime. [Remember: All you need to do is to give a counterexample.] (ii) Show, by giving an example, that if p is not prime, then xP = x mod p can still be true for some integers x with x -1, 0, 1.

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