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 Recognitions: Gold Member Well, a^2-b^2 = (a+b)(a-b), right? Then consider the cases of a and b even, and a and b odd (if only one is even, then the left side is odd and the right side is even, a contradiction). You'll find that (a+b) is even AND (a-b) is ALSO even, and thus the left side is divisible by four, while the right side cannot be divisible by four. Again, contradiction.

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 Quote by Dick The usual trick for problems like this use modular arithmetic. What are the possible values of a^2 mod 4?
I assumed by the title that he was looking for a proof by contradiction, but we're probably working toward the same thing anyway.

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 Quote by Char. Limit I assumed by the title that he was looking for a proof by contradiction, but we're probably working toward the same thing anyway.
Right. That's why I deleted my post. We are talking about the same thing but your wording is more likely what the problem is looking for.

 Thanks. I tried this before but I must have messed up the algebra or something