# Confusion on last step of Hardy's proof on the square root of “a rational number with imperfect square component”'s inability to be a rational number

Here is a copy of Hardy's proof:

For suppose, if possible, that $$p^2/q^2 = m/n$$, $$p$$ having no factor in common with $$q$$, and $$m$$ no factor in common with $$n$$. Then $$np^2 = mq^2$$. Every factor of $$q^2$$ must divide $$np^2$$, and as $$p$$ and $$q$$ have no common factor, every factor of $$q^2$$ must divide $$n$$. Hence $$n = λq^2$$, where $$λ$$ is an integer. But this involves $$m = λp^2$$: and as $$m$$ and $$n$$ have no common factor, $$λ$$ must be unity. Thus $$m = p^2$$, $$n = q^2$$, as was to be proved. In particular it follows, by taking $$n = 1$$, that an integer cannot be the square of a rational number, unless that rational number is itself integral.

I understand the reasoning that gets us to this point, but I don't understand what $$m = p^2$$, $$n = q^2$$ tells us about anything at all related to the original goal of the proof, which is in Hardy's words:

In fact we may go further and say that there is no rational number whose square is $$m/n$$, where $$m/n$$ is any positive fraction in its lowest terms, unless $$m$$ and $$n$$ are both perfect squares.

If you understand it so far (up to end of first yellow box of your post) it has then already been shown that if $$m/n=p^2/q^2$$ then $$m=p^2$$ and $$n=q^2.$$ Hardy in the proof assumes that $$m/n$$ (lowest terms) is the square of a rational. He temporarily calls that rational (before squaring) $$p/q.$$ He may of course assume $$\gcd(p,q)=1, \ \gcd(m,n)=1$$ which he does near the start.
At the end he has shown that $$m=p^2,n=q^2,$$ i.e. the numerator and denominator of $$m/n$$ are both perfect squares, which is what he's trying to prove.
• Wouldn't this just show that $m$ and $n$ can be perfect squares? Why does this show that they have to? – Vityou Nov 20 '18 at 3:41
• I guess you didn't understand Hardy's proof (the part in the first yellow box) well enough. Right after the "Thus:" in that box, he has derived that $m=p^2,n=q^2$ has to be true, i.e. $m,n$ have to be squares. Please add to your question (rather than in another comment) exactly which step(s) in the first yellow box you have trouble with. – coffeemath Nov 20 '18 at 4:42