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I was browsing a set of number theory problems, and I came across this one:

"Prove that the equation a^{2}+b^{2}=c^{2}+3 has infinitely many solutions in integers."

Now, I found out that c must be odd and a and b must be even. So, for some integer n, c=2n+1, so c^{2}+3=4n^{2}+4n+4=4[n^{2}+n+1]. If n is of the form k^{2}-1, then the triple of integers{2n,2[itex]\sqrt{n+1}[/itex],2n+1]} satisfies the equation. Since there are infinitely such n, the equation holds for integers infinitely often.

I thought this was cool.

Mathguy

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# Another interesting number theory tidbit

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