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I have to prove that for every $n>1$, $n^4+4$ gives a non prime number and I thought that I'll have to write it in difference of two squares format. What are the steps to do that and how is its final form? Is that gonna help me prove the original sentence?

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We have $n^4+4=(n^2+2)^2-(2n)^2$. Now use the identity $x^2-y^2=(x+y)(x-y)$.

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$$n^4+4=\left(n^2+2\right)^2-(2n)^2=\left(n^2+2+2n\right)\left(n^2+2-2n\right)$$

More generally, $$n^4+4m^4=\left(n^2+2m^2\right)^2-(2nm)^2=\left(n^2+2m^2+2nm\right)\left(n^2+2m^2-2nm\right)$$

The latter is called Sophie-Germain identity.

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