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# Let a,b,c be positive integers such that a divides b2, b divides c2 and c divides a2. Prove that abc divides (a+b+c)7

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## If a prime p divides a, then p|b2 and hence p|b. This implies that p|c2 and hence p|c. Thus every prime dividing a also divides b and c. By symmetry, this is true for b and 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.) Assume min {x, y, z} = x. Now b|c2 implies that y≤2z; c|a2 implies that z≤2x.We obtain y≤2z≤4x.∴x+y+z≤x+2x+4x=7x. Hence the maximum power of p that divides abc is x+y+z≤7x. Since x is the minimum among x, y, z, px divides a, b, c. ∴px divides a + b + c. This implies that p7x divides (a+b+c)7. Since x+y+z≤7x, it follows that px+y+z divides (a+b+c)7. This is true of any prime p dividing a, b, c. Hence abc divides (a+b+c)7.  Suggest Corrections  0      Similar questions  Related Videos   Conjugate of a Complex Number
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