Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$.

Question

I need help proving the following: Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I know that the smallest $n$ is 8 by testing $n=1,2,3,...$. I need help proving that $n=8$ is the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$.

Answer 1

It is well-known that $x^4+4=\Big(x^2\Big)^2+2^2+\underbrace{2\cdot2x^2-2\cdot2x^2}_0=\Big(x^2+2\Big)^2-4x^2=$

$=\big(x^2-2x+2\big)\big(x^2+2x+2\big)$. However, since $x^2\pm2x+2=\pm1$ has as viable integer

solutions $x=\pm1$, we must extend the search to polynomials of the form $x^2+4a^4$, which

has no such solutions for $a=2\iff n=8$.

Answer 2

Hint: $n$ is not very big. When you find an $n$ that doesn't seem to give any primes, try factoring the polynomial $x^4 + n^2$.

Answer 3

Hint: $$ x^4+4m^4 = (x^4+4x^2m^2+4m^4)-(4x^2m^2)=(x^2+2m^2)^2-(2mx)^2 = (x^2-2mx+2m^2)(x^2+2mx+2m^2).$$