As said below, we only need to consider the odd n's. Let n=2k+1, where k is a positive integer.
n4 + 4n = n4 + 42k+1
Now if we can factor this we are done. This expression has the form a4+4b4, where a = n, b = 2k. But
a4+4b4 = (a2+2ab+2b2)(a2-2ab+2b2)
n4+4(2k)4 = [(n2+2(n)(2k)+2(2k)2]x[(n2-2(n)(2k)+2(2k)2]
Now it remains to check that both factors are not 1. Notice the first term is always bigger than 1 as n>1, k>=1. The second term can be reexpressed as
(n2-2(n)(2k)+2(2k)2 = n2-n2k+1+22k+1
If 2k-n = 2k-2k-1 is positive then we are done. However, simply notice for k>=3, it is indeed positive. So we'll need to check for k=1 and k=2.
But for k = 1 and k=2, or equivalently, n=3 and n=5,
n4 + 4n = 145 and 1649 respectively, which can be factored by 5 and 17 respectively. Therefore we are done.
Edited on August 12, 2004, 3:35 pm
Posted by Bon
on 2004-08-12 15:31:37