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 $x$.
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What do you mean by "find" ,because proving that there exist such an $x$ is difficult and so finding an expression of $n$ is more difficult and if you want a program which can find this element you need to test all elements from $1$ until you find such an element – Elaqqad May 03 '15 at 20:31
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It should be worded differently. I know that the smallest n is 8 by testing the numbers. I needed help proving that 8 is the smallest possible n so that $x^4+n^2$ is not a prime for any $x$. – Dawn AM May 04 '15 at 17:30
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Hint: $n$ is not very big. When you find an $n$ that doesn't seem to give any primes, try factoring the polynomial $x^4 + n^2$.
Robert Israel
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Hint: $$ x^4+4m^4 = (x^4+4x^2m^2+4m^4)-(4x^2m^2)=(x^2+2m^2)^2-(2mx)^2 = (x^2-2mx+2m^2)(x^2+2mx+2m^2).$$
Jack D'Aurizio
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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 $a=2\iff n=8$.
Lucian
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