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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

Evaluate the sum from n = 1 to infinity of n/(n^4 + 4)

Source: Traditional

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