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Below is a direct proof that if $p,q,n$ are positive integers with $\gcd(p,q)=1$ and $p^2=nq^2$ then $q=1$ (so $n=p^2$). I would count that as a direct proof that $$\lbrace n \mid \sqrt{n}\in \mathbb{Q} \rbrace=\lbrace0,1,4,9,16,\cdots\rbrace$$ Given that $\gcd(p,q)=1$ there are integers $s,t$ with $ps+qt=1.$ Cube and regroup to get $p^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1.$ Given that $p^2=nq^2$ we then have $nq^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1$ so that $q$ divides 1. QED
later As Andres points out, it suffices to square. The cubing shows that $\gcd(p,q)=1$ implies $\gcd(p^2,q^2)=1$. Of course if $p,q,n$ are integers and we already know $\frac{p^2}{q^2} \ne n$ then it follows that $|\frac{p^2}{q^2}-n| \ge \frac{1}{q^2}$. I wanted an direct argument that if $p,q,n$ are positive integers with $\gcd(p,q)=1$ and $q \ge 2$ then $|\frac{p^2}{q^2}-n| \ge \frac{1}{q^2}$. I think that could be done but in this situation one wants to keep a proof short.
Below is a direct proof that if $p,q$ p,q,n$are positive integers with$\gcd(p,q)=1$and$p^2=nq^2$then$q=1$(so$n=p^2$). I would count that as a direct proof that $$\lbrace n \mid \sqrt{n}\in \mathbb{Q} \rbrace=\lbrace0,1,4,9,16,\cdots\rbrace$$ Given that$\gcd(p,q)=1$there are integers$s,t$with$ps+qt=1.$Cube and regroup to get$p^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1.$Given that$p^2=nq^2$we then have$nq^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1$so that$q$divides 1. QED 1 Below is a direct proof that if$p,q$are positive integers with$\gcd(p,q)=1$and$p^2=nq^2$then$q=1$(so$n=p^2$). I would count that as a direct proof that $$\lbrace n \mid \sqrt{n}\in \mathbb{Q} \rbrace=\lbrace0,1,4,9,16,\cdots\rbrace$$ Given that$\gcd(p,q)=1$there are integers$s,t$with$ps+qt=1.$Cube and regroup to get$p^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1.$Given that$p^2=nq^2$we then have$nq^2(ps+3qt)s^2+q^2(3ps+qt)t^2=1$so that$q\$ divides 1. QED