# When is $4n^4+1$ prime?

Find all natural numbers $n$ such that $4n^4+1$ is prime. $4n^4+1$ is obviously prime when $n=1$. But can we prove that no other $n$ works?

it is $$4n^4+1=(2n^2)^2-2\cdot 2n^2+1+2\cdot 2n^2=(2n^2+1)^2-4n^2=(2n^2+1-2n)(2n^2-1-2n)$$

• Inside last brackets should be +1 - I guess a little typo error... – usiro Mar 6 '18 at 21:32

This is another one of those questions for which Wolfram|Alpha, factordb.com and Sloane's OEIS can point you to the answer.

In Wolfram|Alpha, try Select[4Range[50]^4 + 1, PrimeQ]. Hmm, just 5. Widen the range: Select[4Range[0, 199]^4 + 1, PrimeQ]. Again, just 5. (We don't need to look at negative $n$ because $n^4$ will be positive anyway).

On factordb, be sure to use *, the parser will probably misunderstand a tacit multiplication operator. So, query 4 * n^4 + 1 and you will see that most of these numbers are divisible by 5.

But not all of them. Quite a few of them are the product of exactly two distinct primes, neither of them 5, like 2501 and 1562501. What's going on here?

By chance, are you in a class and Sophie Germain's identity is coming up? From AoPS:

The Sophie Germain Identity states that $$a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$$

Remember that $1^4 = 1$. Therefore $$4n^4 + 1 = 1^4 + 4n^4 = (1^2 + 2n^2 + 2n)(1^2 + 2n^2 - 2n)$$ $$= (2n^2 - 2n + 1)(2n^2 + 2n + 1).$$

And since the first multiplicand is guaranteed to be greater than 1 for any $n$ with absolute value greater than 1, and $(2n^2 - 2n + 1) < (2n^2 + 2n + 1)$ if $n > 0$, we can conclude that $4n^4 + 1$ is the product of at least two distinct primes, sometimes with repetition, if $n < -1$ or $n > 1$.

The case $n = 0$ is checked easily enough. Although 1 has never been a prime number, it's possible Sophie Germain did not know this. That does not diminish her insight in the least.

Hint:

$$2^{4m+2}n^4+1=(2n^{2m+1}+1)^2-(2^{m+1}n)^2$$

Here $m=0$

and $2n^2+1\pm2n=n^2+(n\pm1)^2$ are $>1$ for $n>1$