$\sqrt{d}\in{\mathbb Z}_+$ or $\sqrt{d}\in {\mathbb R}\setminus{\mathbb Q}$ for every positive integer $d$? I just guess that the following statement is true:

$\sqrt{d}\in{\mathbb Z}_+$ or $\sqrt{d}\in {\mathbb R}\setminus{\mathbb Q}$ for positive integer $d$?

But I don't see a way to deal with it. I think the point is to discuss $dn^2=m^2$ where $n$ and $m$ are positive integers. If I can conclude that $n|m$ then the proof will be done. Help? 
 A: Lemma: If $a|bc$ and $(a,b)=1$ , then $a|c$.
Lemma: If $(m,n)=1$ then $(m^2,n^2)=1$.
Theorem: If $(m,n)=1$ and $dn^2=m^2$, then $d=m^2$.
Proof of theorem only: If $dn^2=m^2$, then $m^2|dn^2$ and therefore $m^2|d$ by lemmas. So $m^2|d$ and $d|m^2$ implies $d=\pm m^2$. But clearly $d$ is positive.
Both of the lemmas can be proven using the fact that if $(a,b)=1$ then $ax+by=1$ has a solution for $x,y\in\mathbb Z$.
A: $dn^2=m^2 $. If $p|d$ prime such that $p^{2n+1}||d$ (such a prime exists) then $p^{2n+1}||m^2$ , which is clearly impossible.
A: More directly, without assuming $m,n$ are coprime, you can try the following steps:


*

*$n\vert m$ if and only if for any prime $p$ and natural number $k$, if $p^k$ divides $n$, then it divides $m$ (by prime factorization).

*Choose arbitrary $p^k$ dividing $n$. Then $p^{2k}$ divides $n^2$ and consequently $m^2$.

*If $p^k$ did not divide $m$, then by uniqueness of factorization, $p^{2k}$ could not divide $m^2$, so $p^k$ divides $m$ and we're done.


In fact, you can prove a similar theorem for arbitrary UFD in the same way: any element of an UFD $R$ has a square root in $R$ if and only if it has one in the quotient field. This is a special case of a more general fact that a unique factorization domain is always integrally closed.
