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I searched a prime of the form $4^n+n^4$ with $n\ge 2$ and did not find one with $n\le 12\ 000$.

  • If $n$ is even, then $4^n+n^4$ is even, so it cannot be prime.
  • If $n$ is odd and not divisible by $5$ , then $4^n+n^4\equiv (-1)+1\equiv 0 \pmod 5$.

    So, $n$ must have the form $10k+5$.

    For $n=35$ and $n=55$, the number $4^n+n^4$ splits into two primes with almost the same size.

    So, is there an obvious reason (like algebraic factors) that there is no prime I am looking for ?

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marked as duplicate by Travis, Shaun, kingW3, Martin Sleziak, Charles Jan 27 '15 at 15:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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If $n$ is even, so will be $4^n+n^4$ and the later will definitely be $>2$ and hence composite

Else

$$4^n+n^4=(2^n)^2+(n^2)^2=(n^2+2^n)^2-2\cdot2^n\cdot n^2$$

$$=(n^2+2^n)^2-(n2^{\frac{n+1}2})^2$$

As $n$ is odd $\iff n+1$ is even $\implies\dfrac{n+1}2$ is an integer

$$4^n+n^4=(n^2+2^n+n2^{\frac{n+1}2})(n^2+2^n-n2^{\frac{n+1}2}) $$

Establish that both factors are $>1$

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  • $\begingroup$ You may want to add that $n$ is odd $\implies\frac{n+1}{2}\in\mathbb{N}$ (and of course, that the expression above is in the form of $a^2-b^2=(a+b)(a-b)$). $\endgroup$ – barak manos Jan 27 '15 at 12:50
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    $\begingroup$ @barakmanos, Please find the edited version $\endgroup$ – lab bhattacharjee Jan 27 '15 at 13:03
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An alternative to @labbhattacharjee’s exemplary response (for the case of odd $n$ only):

(1) $X^4 + 4=(X^2-2X+2)(X^2+2X+2)$; (2) $X^4+4a^4=(X^2-2aX+2a^2)(X^2+2aX+2a^2)$; (3) $X^4+4^{2k+1}=X^4+4\cdot2^{4k}$.

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