Abstract algebra: If p is a prime show that the solutions of X^2= [1]_p in ℤp are precisely [1]_p and [-1]_p?
a) If p is a prime show that the slutions of X^2=[1]_p in ℤ_p are precisely [1]_p and [-1]_p
(for the ring of mod integers mod p is ℤ_p)
b) if p > 2 show that the solutions of X^2=[1]_(p^n) are [1]_(p^n) and [-1]_(p^n)
note : that _ means subscript
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Note that x^2 = [1]_p in Z_p <=> x^2 = 1 (mod p) <=> x^2 - 1 = (x + 1)(x - 1) = 0 (mod p). So either p | (x + 1) or p | (x - 1). Recall that x is in the field Z_p so x is between 0 to p -1 inclusive. Hence 1 =< x + 1 =< p and -1 =< x - 1 =< p -1. Because p is prime it must be that either x + 1 = p or x - 1 = 0. The 1st equality gives us x = p - 1 = -1 (mod p) and the 2nd equality gives us x = 1 = 1 (mod p). We are done!
the only attainable rational aspects are a million and -a million. in case you plug the two genuinely certainly one of them in for x, you will see that neither is a answer. utilising my graphing calculator, i stumbled on that x = 3.6273651 is the only authentic root.