Number theory question - squares

Question

Suppose I have $n,m,t$ positive integers such as $nm$ is a square and $mt$ is a square, how do I prove that $nt$ is also a square?

I have said: $nm=k^2$, $mt=f^2$ so $nt=(kf)^2/m^2=(kf/m)^2$. I need to prove that $kf/m$ is integer. I have said that $m|f^2$, so there are several options - if $m=1$ we are done, if $m|f$ then we are also done. we are left with $m|f^2$, but not $f$. similarly goes for $m|k^2$, and here I am stuck.

Thanks very Much

Answer 1

$(nm)(mt)$ is a perfect square. That is, $(nt)m^2$ is a perfect square. So $nt$ must be a perfect square.

Are you perhaps in need of a lemma that if $A$ is a perfect square and $AB$ is a perfect square, then $B$ is a perfect square? This is pretty easy to show if you use the fundamental theorem of arithmetic (unique factorization into prime powers.)

Answer 2

The rational root $\,kf/m\,$ of $\, x^2\!-nt\,$ is an integer by the Rational Root Test. Thus $\,nt\,$ is a square.