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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 .

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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 px ||a, py|| b and pz||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 px divides a + b + c. This implies that p21x divides (a + b + c)21. since x + y + z ≤ 21x, it follows that px + y + z divides (a + b + c)21. this is true of any prime p dividing, c. hence abc divides (a + b + c)21.

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