Question
If n is positive integer that is larger than 1, show that
n^4+4^n is a composite number
If n is positive integer that is larger than 1, show that
n^4+4^n is a composite number
If n is an even number
Let n = 2k (k>1)
n^4+4^n
= (2k)^4 +4^{2k}
=2^4\cdot k^4+4\cdot 4^{2k-1}
=4(4k^4+4^{2k-1})
Therefore n^4+4^n is a composite number
If n is an odd number
Let n = 2k+1 (k>1)
n^4+4^n
= (n^2)^2 +(2^n)^2
= (n^2+2^n)^2 - 2n^22^n
= (n^2+2^n)^2 - n^22^{n+1}
= (n^2+2^n)^2 - n^22^{2k+2}
= (n^2+2^n)^2 - (n2^{k+})^2
=(n^2+2^n+n2^{k+})(n^2+2^n-n2^{k+})
Therefore
n^4+4^n is composite number when n is odd.
In summary, n^4+4^n is a composite number for both even and odd integers that is larger than 1