Prove: 3/ a(2a^2 + 7)

actually a = 1 does work

a(2*1+7) = 9

a = 2
2(2*4+7) = 30

a = 3
3(2*9+7) obviously works but 75

a=4
4(2*16+7) = 4*39 etc
however I do not know how to do the proof

try recurssion ?

(a+1)(2(a+1)^2+7)
(a+1)(2 a^2 + 4 a + 9)
= 2 a^3 + 6 a^2 + 13 a + 9
subtract original 2a^3 +7a
and get
6 a^2 + 6 a + 9
so
the difference between each successive value of a is divisible by 3

Is that form of proof allowed?

You can also work Mod 3:

a(2 a^2 + 7) = a(-a^2 + 1)

Then, if a = 0, the expression is zero. Else, we have that by Fermat's little theorem that a^2 = 1. So, working Mod 3, the expression is always equal to zero, which means that the original expression (not reduced Mod 3) is always divisible by 3.

Huh?

Yes, you are to prove that 3 divides [(a)(2a^2+7)] probably by using the greatest common divisor.

Sorry, can not help

There must be more. For instance, is a is 1, 3 is not a divisor, if a=0, 3 is not a divisor.

Because is such a small number you don't have to use Fermat's little theorem. You can simply say that Mod 3, a can be 0, 1 or 2. Then we have 1^2 = 1 and 2^2 = 4 = 1. So, if a is not zero, a^2 = 1.

convert

36in.=_ft