Solution to puzzle 36: Composite numbers

We have m = ab = cd, where ab and cd are distinct factorizations, and a, b, c, d 1.

Clearly ab is a multiple of c.
By the Fundamental Theorem of Arithmetic, each prime factor of c occurs in a or b.
Let c = rs, where r is the product of the prime factors of c that occur in a, and s is the product of the prime factors of c that occur in b.
(In the absence of any common prime factors, r or s, or both, may equal 1.)

Then there exist u and v (possibly equal to 1) such that:
a = ru
b = sv

We also have d = ab/c.
Hence d = (ru)(sv)/(rs) = uv.

We can therefore write:

an + bn + cn + dn	= rnun + snvn + rnsn + unvn
	= (rn + vn)(sn + un)

Both factors are greater than 1, for all integers n 0.

Therefore aⁿ + bⁿ + cⁿ + dⁿ is composite.

Source: Composite Number on Eric Weisstein's World of Mathematics

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