If a prime p divides a, then p|b^{4} and hence p|b. This implies that p|c^{4} and hence |c. Thus every prime divides a also divides b and c. By symmetry, this is true for b & c as well. We conclude that a, b, c have the same set of prime divisors.

Let p^{x} ||a, p^{y}|| b and p^{z}||c. (Here we write px || a to mean px ||a and (|px + 1|)/a we may assume min {x, y, z} = x. Now b|c4 implies that y ≤ 4z : c| a^{4} implies that z ≤ 4x. we obtain

y ≤ z ≤ 16x

Thus x + y + z ≤ x + 4x + 16x = 21x Hence the maximum power of p that dividesabc is x + y + z ≤ 21x. Since x is the minimum among x, y, z, px divides a, b, c. hence p^{x }divides a + b + c. This implies that p^{21x} divides (a + b + c)^{21.} since x + y + z ≤ 21x, it follows that p^{x + y + z }divides (a + b + c)^{21}. this is true of any prime p dividing, c. hence abc divides (a + b + c)^{21}.