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.

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.

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

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 $n>1$

Updated October 05, 2017 21:20 PM

Updated September 21, 2017 16:20 PM

Updated September 21, 2017 22:20 PM

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