1. ## Prime Question

For which positive integers n is $\displaystyle n^4 + 4^n$ prime?

I know 1 works. I know it can't be an even n because that would end up even, and thus not prime.

I checked 3, 5, and 7 and they all don't work. I don't know how to prove this as working or not though for bigger odd numbers. I feel like 1 might be the only answer, but not sure how to prove that other numbers do/don't work.

2. Originally Posted by Janu42
For which positive integers n is $\displaystyle n^4 + 4^n$ prime?

I know 1 works. I know it can't be an even n because that would end up even, and thus not prime.

I checked 3, 5, and 7 and they all don't work. I don't know how to prove this as working or not though for bigger odd numbers. I feel like 1 might be the only answer, but not sure how to prove that other numbers do/don't work.
You can also eliminate non-multiples of 5 when you consider mod 5. This leaves only n that are congruent to 5 mod 10.

Not sure how to deal with those.. but the first 200 of them are composite.

Maybe you can factor them. Let f(n) = n^4 + 4^n. There might be a pattern here: (note: f(25) is not given as prime factorization, rather I multiplied some primes to get two factors close together.)

f(5) = 17 * 97
f(15) = 29153 * 36833
f(25) = 33350257 * 33759857
f(35) = 34350564553 * 34368914633

Difference between the two factors shows a pattern:

80 = 2^4 * 5
7680 = 2^9 * 3 * 5
409600 = 2^14 * 5 * 5
18350080 = 2^19 * 5 * 7

3. Since we're dealing with $\displaystyle n$ odd we can factor this way:

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

And for $\displaystyle n>1$ both factors are greater than one - the left factor is

obvious. As for the second one - for the odds $\displaystyle 3, 5$ it can be checked

by hand. For $\displaystyle n\ge 7$ we have

$\displaystyle n<2^\frac{n-1}{2} \Rightarrow 2^\frac{n+1}{2}n<2^n \Rightarrow 2^n-2^\frac{n+1}{2}n > 0 \Rightarrow n^2+2^n-2^\frac{n+1}{2}n > 1$