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I am taking a basic algebra course, and one of the proposed problems asks to prove that $n^4 + 4$ is never a prime number for $n>1$.

I am able to prove it in some particular cases, but I am not able to do it when $n$ is an odd multiple of $5$.

E.L.
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  • If $n$ is an even multiple of $5$, $n$ is a multiple of $10$ so $n^4 + 4$ is even. – Michael Albanese Jan 09 '14 at 14:59
  • See also http://math.stackexchange.com/questions/1121407/show-that-n44-is-not-a-prime-number and http://math.stackexchange.com/questions/581764/for-what-values-of-n-n44-is-composite-number – Martin Sleziak Jan 27 '15 at 09:10

3 Answers3

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$n^4 + 4 = (2 - 2 n + n^2) (2 + 2 n + n^2)$

Xeing
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Hint $\ $ Completing the square leads to a difference of squares

$$\begin{eqnarray} &&\ \, n^4 + 2^2\\ \,&=&\, (n^2\!+2)^2\!-(2n)^2\\ \,&=&\, (n^2\!+2\ -\ 2n)\ (n^2\!+2\ +\ 2n)\end{eqnarray}$$

Community
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Bill Dubuque
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More generalized: $a^4 + 4b^4 = a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 = (a^2 + 2b^2)^2 - (2ab)^2 = (a^2+2b^2-2ab)(a^2+2b^2+2ab)$

user134489
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