As said below, we only need to consider the odd n's.  Let n=2k+1, where k is a positive integer.

Then

n⁴ + 4ⁿ = n⁴ + 4^2k+1
= n⁴+4(2^k)⁴

Now if we can factor this we are done.  This expression has the form a⁴+4b⁴, where a = n, b = 2^k.  But
a⁴+4b⁴ = (a²+2ab+2b²)(a²-2ab+2b²)

Therefore,

n⁴+4(2^k)⁴ = [(n²+2(n)(2^k)+2(2^k)²]x[(n²-2(n)(2^k)+2(2^k)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)(2^k)+2(2^k)² = n²-n2^k+1+2^2k+1
= n²+2^k+1(2^k-n)

If 2^k-n = 2^k-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