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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 $a<b$. 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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2 Answers 2

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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That's what I was originally doing. I know $a \text{ and } b\equiv\pm1 \text{ or }\pm3 \bmod{8}$ $a^2+2 \equiv 3\bmod{8}$ and $b^2+4\equiv -3\bmod{8}$. I don't see any relation between what kinds of primes must divide the two though –  Mike M. Nov 9 '12 at 5:24
    
Ah, I see that the primes dividing $a^2+2$ must be $1 or 3 \bmod{8}$ and the primes dividing $a^2+4$ must be $-1 or -3 \bmod{8}$. Is that all that needs to be done seeing as the prime factors of $a^2+2$ will never be any of the prime factors of $b^2 +4$? I see this by calculation only though. What would be a good way to show those are the only possible prime divisors? –  Mike M. Nov 9 '12 at 5:44
    
I think I got it using the Legendre symbol and your hint! Realized I made a mistake in my earlier post regarding primes dividing $b^2 +4$, but that was fixed in the proof. Thank you for your help! –  Mike M. Nov 9 '12 at 6:31
    
I'm glad you worked it out. Meanwhile, it looks like @dinoboy got there, too. –  Gerry Myerson Nov 9 '12 at 11:29

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