Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far:
$a$ is odd, so $a = 2k + 1$ for some integer $k$.
Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$  
$= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $
$=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k + 32$
$=16k^4 + 32k^3 + 64k^2 + 48k + 32$
But this isn't a multiple of 32, at most I could say $(a^2 + 3)(a^2 + 7) = 16b$ for some integer $b$

Notice
$$16k^4 +48k = 16k(k^3+3).$$
If $k$ is even, we are done. If $k$ is odd, then $k^3 +3$ is even, and we are done.
