Prove that \mathbf{n^4 + 4^{n}} is composite for all values of n greater than 1.

Discussion:

Teacher: This problem uses an identity that has a fancy name: Sophie Germain identity. But what’s in a name after all.

Clearly if n is even the expression is composite as it is divisible by 2. We have to check what happens when n is odd.

Student: I remember Sophie Germain’s identity. It says that a^4 + 4b^4 can be further factorized. As you hinted we can use it here.

Suppose n = 2k +1 (for some k).

So n^4 + 4^{2k+1} = n^4 + 4\cdot (4^k)^2 = (n^2 + 2\cdot 4^k)^2 - 2 \times n^2 \cdot 2 \cdot 4^k = (n^2 + 2 \cdot 4^k)^2 - (2 \cdot n \cdot 2^k )^2

Thus we can use a^2 - b^2 = (a+b)(a-b) identity.

So n^4 + 4^n = (n^2 + 2 \cdot 4^k + 2n \cdot 2^k )(n^2 + 2 \cdot 4^k - 2n \cdot 2^k )