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^210n+2) (25n^2+10n+2), and the same can also be done with all the others:
(n^22n+2) (n^2+2n+2) =n^4+4
4(2n^22n+1) (2n^2+2n+1) =(2n)^4+4
(9n^26n+2) (9n^2+6n+2) =(3n)^4+4
4(8n^24n+1) (8n^2+4n+1) =(4n)^4+4
(25n^210n+2) (25n^2+10n+2) = (5n)^4+4
Edited on July 12, 2016, 10:35 am

Posted by broll
on 20160712 10:33:01 