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