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# Sophie Germain (B.Stat 2006, subjective problem 3)

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

May 8, 2015