# $n^4 + 4^n$ is a not a prime

by Tamaghna Chaudhuri   Last Updated August 14, 2019 06:20 AM

Prove that $$n^4 + 4^n$$ is not a prime for all $$n > 1$$ and $$n \in \mathbb{N}$$.

This question appeared in the undergrad entrance exam of the Indian Statistical institute.

When $$n$$ is even the proof is simple. For $$n = 2m+1$$ I am utterly stuck.

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Let $$n=2k+1$$ with $$k\geq 1$$, then $$n^4+4^n=n^4+4 \cdot 4^{2k}=n^4+4\cdot (2^k)^4=(n^2+2\cdot 2^{2k}+2^{k+1}n)(n^2+2\cdot 2^{2k}-2^{k+1}n).$$ Thus $$n^4+4^n$$ can be factored into non-trivial factors, when $$n$$ is odd.

Anurag A
August 14, 2019 05:42 AM

Remember doing this as an Exercise from Ivan Niven's Number Theory book. I think this follows from an identity of Sophie Germain

crskhr
August 14, 2019 05:46 AM

Firstly, I’ll show that we can factorise $$x^4+4y^4$$. $$x^4+4y^4=x^4-4x^2y^2+4y^4-4x^2y^2= \left(x^2+2y^2\right)^2-\left(2xy\right)^2= \left(x^2+2xy+2y^2\right)\left(x^2-2xy+2y^2\right)$$

Back to the question, when $$n=2m+1$$ that $$m$$ is a positive integer, we can substitute $$x=n$$ and $$y=2^m$$ into the polynomial above, we’ll get $$n^4+4\times\left(2^m\right)^4=n^4+4\times4^{2m}=n^4+4^{2m+1}=n^4+4^n=\left(n^2+2^{m+1}n+2^{2m+1}\right)\left(n^2-2^{m+1}n+2^{2m+1}\right)$$ Therefore, $$n^4+4^n$$ is not a prime for all odd number $$n>1$$

Isaac YIU Math Studio
August 14, 2019 05:56 AM

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