Divisibility theory help

Question

If $a$ is odd, show that $32 \mid (a^2 + 3)(a^2 + 7)$

Since $a$ is odd, I let $a = 2b + 1$ and did the expansion to get $16\mid [(b^2 + b +1)(b^2 + b + 2)]$, but I was unable to continue from here. Is it the correct method ?

Can someone also please teach me how to show that

$3 \mid a(2a^2 + 7) $

for any arbitrary integer $a$?

Answer 1

For the second part, that is, $3 \, | \, a(2a^2 + 7)$:

We have that $a^3 \equiv a \pmod{3}$, by Fermat. Hence $2a^3 + 7a \equiv 2a + 7a \pmod{3}$.

But $2a + 7a \equiv 9a \equiv 0\pmod{3}$.

Answer 2

Hint: Consider the congruence class of $a^2+3$ modulo $4$ and the congruence class of $a^2+7$ modulo $8$.

Answer 3

Hint: If $a$ is odd, then $a^2=8m+1$.

Proof: Let $a=4k\pm 1$. Then $(4k\pm 1)^2=16k^2\pm 8k+1$.

For the second one: if $a$ is divisible by $3$, we're done.

Otherwise $2(3k\pm 1)^2+7=18k^2\pm12k+9$ is divisible by $3$.

Answer 4

Let $a=2b+1$. Then $a^2+3=4b^2+4b+4$ and $a^2+7=4b^2+4b+8$. Both of these terms are divisible by 4, so $(a^2+3)(a^2+7)=16(b^2+b+1)(b^2+b+2)$.

Now note that exactly one of $b^2+b+1$ and $b^2+b+2$ must be even since they have difference one (in fact it must be the second term) so therefore

$(a^2+3)(a^2+7)=32(b^2+b+1)\dfrac{b^2+b+2}{2}$, so 32| $(a^2+3)(a^2+7)$.

For the second part, consider the statement modulo 3.

If a=0 mod 3, then we are done (a=0 mod 3 means 3|a so in particular divides any multiple). Otherwise $a^2=1$ mod 3, hence $2a^2+7 = 2+1=0$ mod 3 so 3 divides the second factor and we're done.

Answer 5

$$(a^2+3)(a^2+7)=(a^2-1+4)(a^2-1+8)=\{(a-1)(a+1)\}^2+12(a-1)(a+1)+32$$

Now as $a$ is odd, one of $a-1,a+1$ is divisible by $4$ and the other is by $2,$ not by $4$

$\implies 4\cdot2$ divides $(a-1)(a+1)$