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.

a^{n} + b^{n} + c^{n} + d^{n} | = r^{n}u^{n} + s^{n}v^{n} + r^{n}s^{n} + u^{n}v^{n} |

= (r^{n} + v^{n})(s^{n} + u^{n}) |

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

Therefore a^{n} + b^{n} + c^{n} + d^{n} is composite.

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