If n is an integer, show that n⁴ + 4ⁿ is never a prime for n>1.

If n is even, then 2 divides n⁴ + 4ⁿ.
Clearly (n⁴ + 4ⁿ)/2 > 2, for n > 1.
This establishes the result for even n.

If n is odd, we will write n⁴ + 4ⁿ as the difference of two squares of integers, and hence obtain a factorization.  Setting n = 2m + 1, we have

<table><tbody><tr class="h"><td class="u">n⁴ + 4^2m+1</td><td> = (n²)² + (2^2m+1)²</td></tr> <tr><td> </td><td> = (n² + 2^2m+1)² − 2^2m+2n²</td></tr><tr><td> </td><td> = (n² + 2^2m+1 + 2^m+1n)(n² + 2^2m+1 − 2^m+1n)</td></tr></tbody></table>

We must now show that both factors are greater than one.

Clearly n² + 2^2m+1 + 2^m+1n > 1, for n > 1.

<table><tbody><tr class="h"><td class="u">Next consider f(n,m)</td><td> = n² + 2^2m+1 − 2^m+1n</td></tr><tr><td> </td><td> = (n − 2^m)² + 2^2m</td></tr></tbody></table>

For n > 1 (m > 0), 2^2m > 1; hence f(n,m) > 1 for all odd n > 1.

Therefore n⁴ + 4ⁿ is composite for all integers n > 1.

Posted by Danish Ahmed Khan on 2012-10-24 14:42:01