As said below, we only need to consider the odd n's. Let n=2k+1, where k is a positive integer.
Then
n^{4} + 4^{n} = n^{4} + 4^{2k+1}= n^{4}+4(2^{k})^{4}
Now if we can factor this we are done. This expression has the form a^{4}+4b^{4}, where a = n, b = 2^{k}. But
a^{4}+4b^{4} = (a^{2}+2ab+2b^{2})(a^{2}2ab+2b^{2})
Therefore,
n^{4}+4(2^{k})^{4} = [(n^{2}+2(n)(2^{k})+2(2^{k})^{2}]x[(n^{2}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
(n^{2}2(n)(2^{k})+2(2^{k})^{2} = n^{2}n2^{k+1}+2^{2k+1}= n^{2}+2^{k+1}(2^{k}n)
If 2^{k}n = 2^{k}2k1 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,
n^{4} + 4^{n} = 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 20040812 15:31:37 