Question

Let a,b,c be positive integers such that a divides b^2,b divides c^2,c divided a^2 prove that abc divides (a+b+c)^7.


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Answer 1

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Answer 2

Answer:

 

If a prime p divides a, then p∣b  2  and hence p∣b. This implies that p∣c  2

 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 p  x ∣∣a,p  y ∣∣b and pz∣∣c. (Here we write p  x ∣∣a to mean px∣a and p  x+1  ∣/a.)

Assume min {x, y, z}  = x.

Now b∣c  2   implies that y≤2z; c∣a  2   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, p  

x divides a, b, c.

∴p  x divides a + b + c.

This implies that p  7x divides (a+b+c)  7  .

Since x+y+z≤7x, it follows that p  

x+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  .