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I was wondering whether the following theorem was already established, or if it was something novel. I couldn't find it in my abstract algebra textbook and a Google search was not productive ("a list of congruence theorems" just turns up elementary geometry theorems on triangle congruences.)

ThmFor any n in ℕ, if a+b=n, then a^{2}≡ b^{2}(mod n).

Proof: Let a+b=n. Then a=b-n.

a^{2}= (b-n)^{2}= b^{2}- 2bn + n^{2}= b^{2}- n*(2b)+n) ≡ b^{2}(mod n)

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# A Theorem on Squares and Congruences

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