Start with small values, mod 5. Unless a is divisible by 5, the expression is worth 0, mod 5.

For a divisible by 5, small values are:

629 = 17*37, 10004 = 2^2*41*61, 50629 = 197*257,... etc. where the factors can be separated into 2 parts with a regularly increasing difference of 20,40,60,...etc.

With that hint it is quite easy to find a factorisation of (5n)^4+4 = (25n^2-10n+2) (25n^2+10n+2), and the same can also be done with all the others:

(n^2-2n+2) (n^2+2n+2) =n^4+4   
4(2n^2-2n+1) (2n^2+2n+1) =(2n)^4+4   
(9n^2-6n+2) (9n^2+6n+2) =(3n)^4+4   
4(8n^2-4n+1) (8n^2+4n+1) =(4n)^4+4   
(25n^2-10n+2) (25n^2+10n+2) = (5n)^4+4   

Edited on July 12, 2016, 10:35 am

Posted by broll on 2016-07-12 10:33:01