middleCmusic
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Hey guys --
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.)
Thm For any n in ℕ, if a+b=n, then a2 ≡ b2 (mod n).
Proof: Let a+b=n. Then a=b-n.
a2 = (b-n)2 = b2 - 2bn + n2 = b2 - n*(2b)+n) ≡ b2 (mod n)
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.)
Thm For any n in ℕ, if a+b=n, then a2 ≡ b2 (mod n).
Proof: Let a+b=n. Then a=b-n.
a2 = (b-n)2 = b2 - 2bn + n2 = b2 - n*(2b)+n) ≡ b2 (mod n)