Prove that there are infinitely many natural numbers $a$ with a specific property 
Prove that there are infinitely many natural numbers $a$ with the
  following property: the number $n^4 + a$ is not prime for any
  natural number $n$.


I only need a hint, how to start with.
 A: $(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$
We would like this to be of the form $x^4+w$, so we need $a=-c$.
So the product is equal to: $x^4+(b+d-a^2)x^2+(ad-ab)x+bd$, so we need $b=d$ or $a=0$. But if $a=0$ we'll need $b=-d$ later on, and thatwould just yield $x^4-d^2$.
So the product is equal to: $x^4+(2b-a^2)x^2+bd$.
So the polynomial $(x^2+ax+b)(x^2-ax+b)=x^4+(2b-a^2)x^2+b^2$ works whenever $2b=a^2$.
Therefore the polynomial we want is:
$(x^2+2ax+2a^2)(x^2-2ax+2a^2)=x^4+4a^4$
We just have to find the values of $a$ so that each of the factors never takes on the value $1$ or $-1$ when $x$ is an integer.
So we find the discriminants of $x^2+2ax+2a^2\pm1$ and $x^2-2ax+2a^2\pm1$.
They take on values $4a^2-4(2a^2+1)=4(-1-a^2)$ and $4a^2-4(2a^2-1)=4(1-a^2)$.
Clearly both of these discriminants are negative if $a>1$.
Therefore $x^4+4a^4$ is not an integer for any $x\in\mathbb Z$ and $a\geq 2\in \mathbb Z$
A: Hint:-
Let $a=4b^4$
So,the given expression becomes $a^4+4b^4$.It is never a prime because it can be split into two distinct factors.
[How?-Use Sophie Germains Identity to factorise.]
