1

The expression $n^4 + \,4^n$ is known to be prime only for $n = 1$ corresponding to the prime $5$. The usual proof makes use of the Sophie Germain identity $$a^4 + \,4b^4 ≡ \left\{(a - b)^2 +\, b^2\right\}\left\{(a + b)^2 +\, b^2\right\}$$ since $n^4 + \,4^n$ is odd only for odd $n$ in which case $4^n$ has the form $4\left(2^k\right)^4$ and then both factors $\left(n ± 2^k\right)^2 + \,4^k > 1$ for $ k > 0 $. We observe, on the other hand, that for all $n$ ending in $1, 3, 7, 9$ expression ends in $5$ and obviously cannot be prime. I was wondering if a similar line of thought could dispose of the case when $n$ ends in $5$, but here I can't proceed further beyond the fact that they generate numbers ending in $49$ and primes do end in $49$ and hence doesn't preclude the existence of a prime. Can we somehow obviate recourse to Sophie Germain identity (or the like) to prove the expression is composite also in the case $n$ ending in $5$ ?

Beedassy Lekraj
  • 121
  • 4
    In the end, the Sophie Germain identity is just the distributive law - what could be simpler? – Hagen von Eitzen Nov 13 '20 at 12:58
  • 2
    $n^4 + 4^n=1073792449 = 29153\cdot 36833$ when $n=15$. It seems unlikely that this number can be proved composite using elementary congruences. – lhf Nov 13 '20 at 13:53

1 Answers1

1

In general for non negative $a,b$ $$a^2+b^2 = (a -\sqrt{2ab}+b)(a +\sqrt{2ab}+b)$$

so you can say $$n^4 + 4^n = (n^2 - 2^{\frac{(n+1)}2}n +2^n)(n^2 + 2^{\frac{(n+1)}2}n +2^n)$$

which is the product of two integers when $n$ is odd,

and both those integers are greater than $1$ when $n\ge 3$.

Numbers ending in $5$ are odd and greater than $3$.

Henry
  • 157,058
  • Thanks but if you read carefully my request you'll find this is exactly what I'm trying to avoid !! – Beedassy Lekraj Nov 13 '20 at 13:37
  • I have edited my answer to use a factorisation of $a^2+b^2$. So it is not the Sophie Germain identity though you can derive the Sophie Germain identity from it – Henry Nov 13 '20 at 13:42
  • The Sophie Germain identity is straight forward from x² - y² = (x - y)(x + y) with x = a² + 2b², y = 2ab. – Beedassy Lekraj Nov 13 '20 at 14:21