Prove that $32\mid(a^2+3)(a^2+7)$

Question

If $a$ be an odd integer prove that $32\mid(a^2+3)(a^2+7)$

These are not three or two consecutive integers so they are not divisible by $3$ or $2$ also mathematical induction can not be applied. Any idea how to solve it

Answer 1

If $a$ is an odd integer, then start with $a = 2k+1 \implies a^2 = 4k(k+1)+1...$

Answer 2

$a=2n+1 \Rightarrow a^2=4n^2+4n+1$. Then: $$(a^2+3)(a^2+7)=(4n^2+4n+4)(4n^2+4n+8)=16(n^2+n+1)(n^2+n+2).$$ The numbers in the brackets are two consecutive numbers, so one of them must be even.

Answer 3

Use mod $4$, meaning if $a = 1 \pmod 4\implies a^2+3= 0\pmod 4, a^2+7 = 0\pmod 8\implies (a^2+3)(a^2+7)= 0 \pmod {32}$ . If $a = 3\pmod 4\implies a^2+3 = 0\pmod 4, a^2+7 = 0\pmod 8\implies (a^2+3)(a^2+7) = 0\pmod {32}$ . Either case gives the answer.