Problems & Puzzles: Puzzles

Puzzle 321. Primes type n^4+4

Faride Firoozbakht asks: find a prime greater than 5 of the form n^4 +4 (or prove there is no such prime). 


The same solution came from many puzzlers: Bill Murphy, Adam Stinchcombe, Dan Dima, J. K. Andersen, J. C. Rosa, Jean Brette, Sebastian Martin Ruiz, Salvatore Ingala, Luis Rodríguez, Salvatore Tringali, Rudolph Kjnzek & Ken Wilke:

n^4+4 = (n^2-2n+2) * (n^2+2n+2).
This product can only be prime if one of the factors is +/-1.
The only such prime is 5 for n=1.

 J. C. Rosa pointed out that:

'...Thus 'all number of the form n^4+4 , except 5, is the product of two integers' (Sophie Germain , 1776-1831). I have found this result in "Theory of Numbers " from Edouard Lucas 1891 , page 131'

J. C. Andersen added:

n^4+4 = ((n-1)^2+1) * ((n+1)^2+1) can at best be a semiprime for n>1.
The semiprime problem is finding x^2+1 "twins" for 2 consecutive even x.
It has not been proved there are infinitely many x^2+1 primes,
but there are probably infinitely many semiprimes n^4+4:
n = 3, 5, 15, 25, 55, 125, 205, 385, 465, 635, 645, 715, 1095, ...

PrimeForm/GW found n = 8*1201#/5*(3391948*(4*1201#/5+1)+1)+1.
(n-1)^2+1 and (n+1)^2+1 are 2028-digit primes, giving a 4055-digit semiprime n^4+4.
The form of n was chosen so PrimeForm/GW could prove both primes easily.
The BLS proofs used p = 4*1201#/5+1 is a prime dividing n+1, and p-1 divides n-1.


Nota Bene: This puzzle was not a joke neither a hoax, even if it resulted very simple to solve. As a matter of fact Faride stated this result in the frame of the Puzzle 225 of my pages more than one year ago. But I forgot this frame and I rescued only the question.


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