# Proof about an integer being a perfect square.

1. Jan 11, 2013

### cragar

1. The problem statement, all variables and given/known data
m and n are positive integers with m,n≥2
where $m^2=kn^2$
3. The attempt at a solution
we know that all prime factors of m have an even amount , their are no prime factors that
are repeated an odd number of times. The same goes for n.
if k is not a perfect square then it will have an odd number of prime factors.
and then when we multiply in to n it will still be an odd number, but
m has only an even number, so k has to the square of an integer so both sides have even number of each prime factor.

2. Jan 11, 2013

### Staff: Mentor

6=2*3 has an even number of prime factors.
But it has at least one prime factor where the exponent is odd.

What do you want to prove? That k has to be a perfect square? You can do it that way.

If you can use real numbers in your proof: $a^2=k$ with rational a and an integer k has a solution only if k is a perfect square.

3. Jan 11, 2013

### cragar

what I meant to say is that there would exists a prime factor
that has at least an odd exponent