Solution to puzzle 43: Sum of two powers
Show that n4 + 4n is composite for all integers 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
| n4 + 42m+1 | = (n2)2 + (22m+1)2 |
| = (n2 + 22m+1)2 − 22m+2n2 | |
| = (n2 + 22m+1 + 2m+1n)(n2 + 22m+1 − 2m+1n) |
We must now show that both factors are greater than one.
Clearly n2 + 22m+1 + 2m+1n > 1, for n > 1.
| Next consider f(n,m) | = n2 + 22m+1 − 2m+1n |
| = (n − 2m)2 + 22m |
For n > 1 (m > 0), 22m > 1; hence f(n,m) > 1 for all odd n > 1.
Therefore n4 + 4n is composite for all integers n > 1.
An infinite sum
Source: Traditional