4
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Inside last brackets should be +1 - I guess a little typo error... $\endgroup$ – usiro Mar 6 '18 at 21:32
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $n>1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.