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I need to show that there do not exist integers $a$ and $b$, both odd, for which $a^2+2$|$b^2+4$.

I have broken it into cases of $a>b$, $a=b$, and $a. The first two cases seem obvious, but I'm having difficulty figuring out where to go on the third case, assuming this is a reasonable approach. I have the feeling the legendre symbol might be of good use?

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Note that if $a$ is odd, we have $a^2 + 2 \equiv 3 \pmod{4}$. This implies some prime $3 \pmod{4}$ divides $a^2 + 2$. However, it is well-known no prime $3 \pmod{4}$ divides the some of two squares such as $b^2 + 2^2$ so the result follows because if $a,b$ existed some prime $3 \pmod{4}$ would have to divide $b^2 + 4$.

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Work modulo 8 --- think about what kinds of primes can/must divide $a^2+2$, and what kind $b^2+4$.

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    I'm glad you worked it out. Meanwhile, it looks like @dinoboy got there, too.2012-11-09