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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.

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  • $\begingroup$ Let $a = n{}{}{}{}$? $\endgroup$ – ÍgjøgnumMeg Sep 25 '16 at 16:37
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    $\begingroup$ @Ed_4434 $a$ doesn't depend on $n$ $\endgroup$ – user261263 Sep 25 '16 at 16:38
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$(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$

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 $a\geq 2\in \mathbb Z$

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  • $\begingroup$ "$a \ge 2 \in \mathbb{Z}$" bothers me a lot. Good answer though. $\endgroup$ – 6005 Sep 25 '16 at 21:27
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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.]

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