Question

Let a, b, c be positive integers such that a divides b⁴ , b divides c⁴ and c divides a⁴. Prove that abc divides (a + b + c)²¹ .

Answer 1

If a prime p  divides a, then p|b⁴ and hence p|b. This implies that p|c⁴ 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| a⁴ 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.