**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 \(n^4 + 4^n = (n^2 + 2 \cdot 4^k + 2n \cdot 2^k )(n^2 + 2 \cdot 4^k – 2n \cdot 2^k )\)

## 1 comment