1) Suppose n belongs to Z.
(a) Prove that if n is congruent to 2 (mod 4), then n is not a difference of two squares.
(b) Prove that if n is not congruent to 2 (mod 4), then n is a difference of two squares.
3) Let n = 3^(t-1). Show that 2^n is congruent to -1 (mod 3^t).
5) Let p be an odd prime, and n = 2p. Show that a^(n-1) is congruent to a (mod n) for any integer a.
Please refer to the attached image for questions with proper symbol notations.© BrainMass Inc. brainmass.com October 10, 2019, 2:03 am ad1c9bdddf
Please see the attachment for full solution.
(a) Since n=2 (mod 4), then n is an even number. If n is the difference of two squares, say n = a^2 - b^2, then either both a and b are even, or both a and b are odd. If a = 2r, b = 2s, then n = a^2 - b^2 = 4(r^2 - s^2 = 0 (mod 4), we get a contradiction; if a = 2r + 1, b = 2s + 1, then we have n = a^2 + b^2 = 4r^2 + 4r - 4s^2 - 4s = 0 (mod 4).
For both cases we get a contradiction. Thus can not be the difference of two squares.
(b) If ...
This posting exemplifies congruency. Step by step calculations are provided.