Proving a and b are squares of an integer

[gottfried]

Homework Statement

If a, b and c are positive integers with (a,b) = 1 and c² = ab, show that each of a and b is the square of an integer.


I've been staring at this problem for a while now and I've got not clues. I don't see the relation between a and b being relatively prime and the rest of the problem. Any hints would be appreciated

 

[voko]

Can you prove that c does not divide either a or b?

And I think it should be required that a, b, c > 1.

 

[uart]

I've been staring at this problem for a while now and I've got not clues. I don't see the relation between a and b being relatively prime and the rest of the problem. Any hints would be appreciated

All the prime factors of a square (eg c^2) have an even multiplicity. Relative prime "a" and "b" means that they have no prime factors in common. Does that give you a hint?

 

[gottfried]

Yes I think so.

Suppose c divides a and b.
Then a=c.m and b=c.n it follows that c²=².m.n and a=b=c. Obviously this can't be since (a,b)=1

Suppose c divides a but not b.
Then c²=c.m.b and c=m.b. Therefore b|c and c|a implying b|a which clearly can't be since (a,b)=1. This also implies that c can't divide b and not a.

Is that a sufficient proof?

Thanks for the hint but I'm still stuck.
Does it follow from c=(a.b)/c and since c doesn't divide a or b that c must divide the product and some how use that fact.

 

[jgens]

If you can use the fact that the integers are a unique factorization domain, then the proof is pretty much trivial, as uart suggested. Otherwise, consider both the greatest divisor of c that also divides a and the greatest divisor of c that also divides b.

 

[voko]

If c does not divide a and b separately, then c must equal AB, where A divides a and B divides b. What does that imply?