# Primes of form $\frac {(4^m+1)}{5}$

What is known about primes of the form

$$p=\frac{4^m+1}{5}$$

where m is an odd positive integer? Is $13$ the only such prime?

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Checked up through m=129; I found no primes other than m=3. –  imallett Nov 16 '12 at 2:11

EDDDDIITTTT: note $$x^4 + 4 y^4 = (x^2 - 2 x y + 2 y^2) (x^2 + 2 x y + 2 y^2)$$ which is what I noticed. Among other things, this is the statement that the fourth roots of $-4$ are $1+i, 1-i, -1+i, -1-i.$ The particular case where $y$ is a power of $2$ and $x=1$ is attributed to Aurifeuille, a name that refers to a person rather than a country. However, I maintain that it's a good name for a country. Well, Aurifeuillia is a good name for a country.

ORIGINAL: Yes.

Take odd $m=2n+1.$ Then $$p = \left( 1 + 4 \cdot (2^n)^4 \right)/5,$$ $$p = (1 - 2^{n+1} + 2^{2n+1}) (1 + 2^{n+1} + 2^{2n+1}) / 5.$$

For $n \geq 2$ both factors are larger than $5.$

Note that all prime factors $q$ of $p$ must satisfy $q \equiv 1 \pmod 4.$ Indeed, if $r$ is prime and $r \equiv 3 \pmod 4,$ and finally $x^2 + y^2 \equiv 0 \pmod r,$ then $x,y \equiv 0 \pmod r$ and $x^2 + y^2 \equiv 0 \pmod {r^2}.$ In our case we have this $y=1$ so no such $r$ can be a factor.

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This is known as an Aurifeuillian factorization. –  Josh B. Nov 16 '12 at 2:41
@JoshB., I did visit Europe in 1980 but I never made it to Aurifeuillia. Note that $$x^4 + 4 y^4 = (x^2 - 2 x y + 2 y^2) (x^2 + 2 x y + 2 y^2)$$ –  Will Jagy Nov 16 '12 at 3:02
Thank you, I had seen that factorisation before but missed it here. –  Sean Gomes Nov 16 '12 at 3:27
@WillJagy: The name seems to be "in honor of the French mathematician [Léon François] Antoine Aurifeuille (1822-1882)". –  ShreevatsaR Nov 16 '12 at 8:56