Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$.

Question

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 $b$

Answer 1

Rewrite your first step as

$$4(k^2 + k + 1) \cdot 4(k^2 + k + 2) = 16 (k^2 + k + 1)(k^2 + k + 2)$$

Now notice that $k^2 + k + 1$ and $k^2 + k + 2$ have opposite parities....

Answer 2

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.

Answer 3

Use modular arithmetic (this is not the most simple way but it also works very well for similar, more complicated problems). Notice that for any odd $a$ we have $a^2 \equiv 1 \ \text{(mod 4)}$ and $a^2 \equiv 1 \ \text{(mod 8)}$. So $a^2 + 3 \equiv 0\ \text{(mod 4)}$, which means that $a^2 + 3$ is divisable by $4$, and $a^2 + 7 \equiv 0\ \text{(mod 8)}$ (idem dito). So the product is divisible by 32.

Answer 4

insert $a=2b+1$ and you get

$((2b+1)^2 + 3)((2b+1)^2 + 7) /32= 1/2 (b^2+b+1) (b^2+b+2)$

either $(b^2+b+1)$ or $(b^2+b+2)$ will be even so the RHS is an integer for all b.