The number $n^4 + 4$ is never prime for $n>1$

elementary-number-theory divisibility

Solution 1:

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

Solution 2:

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

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

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