# Thread: Find all prime numbers

1. ## Find all prime numbers

Find all prime numbers $K$

and all positive integers $N$

such that

$K = N^4+4.$

2. The only solution is $N=1,K=5$.

$N$ clearly has to be odd. If $N>1$, then $N$ must also be divisible by 5; if it isn’t, $N^4+4$ would be divisible by 5 by Fermat’s little theorem.

Hence if $N>1$, let $N=5(2k-1)$, $k\ge1$.

Then $N^4+4=625(2k-1)^4+4=\left[4(5k-2)^2+1\right]\left[4(5k-3)^2+1\right]$ is a product of two integers both greater than 1 and hence cannot be prime.

3. ## Wrong!!!

We get only 1 as the solution!!!!!

4. Yes, that's what JaneBennet has proven. N = 1 is the only solution yielding K = 5.

5. ## Another variant!

Another variant of the problem is this : Find all n such that n^4+4^n is a prime!

6. Originally Posted by manjil
Another variant of the problem is this : Find all n such that n^4+4^n is a prime!
See This

7. Originally Posted by manjil
We get only 1 as the solution!!!!!
Not even wrong, a solution consists of a pair a positive integer N and a prime K. As Miss Bennet demonstrates there is indeed only one solution and it is N=1, K=5

RonL

8. Originally Posted by perash
Find all prime numbers $K$

and all positive integers $N$

such that

$K = N^4+4.$
I found a better solution.

If $N$ is even, then $N^4+4$ is an even number greater than 2, so it is not prime.

If $N$ is odd, let $N=2m+1$, $m\ge0$.

Then $N^4+4=(2m+1)^4+4=\left(4m^2+1\right)\left(4(m+1)^2 +1\right)$.

If $m>0$, both the factors are greater than 1 and so $N^4+4$ is not prime. Hence the only solution is when $m=0$, i.e. $N=1$.