If p is an odd prime, what are the two solutions to x^2 == 4 (mod p) in the set {1, 2, 3, ..., p-1}?
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Originally Posted by tarheelborn If p is an odd prime, what are the two solutions to x^2 == 4 (mod p) in the set {1, 2, 3, ..., p-1}? If $\displaystyle p $ is odd, then $\displaystyle 2^2 \equiv 4 \mod{p} $ and $\displaystyle (-2)^2\equiv(p-2)^2\equiv4\mod{p} $. So our solutions are $\displaystyle x=2,p-2 $.
What if p=3? Isn't that different?
Originally Posted by tarheelborn What if p=3? Isn't that different? Nope, $\displaystyle 2,3-2=1 $ are solutions.
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