Give two integers $m$ and $n$ such that $n^2$ is a multiple of $m$, $n$ is not a multiple of $m$ and $n > m$ Give two integers $m$ and $n$ such that $n^2$ is a multiple of $m$, $n$ is not a multiple of $m$ and $n &gt; m.$


I think that such two integers don't exist, but even if that were true, I'm having trouble with proving that. I have found clear examples when I neglect the last condition ($n &gt; m$).

Any help on this problem would be much appreciated. The observation that I did make is the fact that since $n^2$ is a multiple of $m$ we can write it as $k * m$, which means that $k$ has to be larger than $n$.

I think that could be a starting point of some sort, but I wasn't capable of going any further.
In general pick $n=pq$, where $p$ is a prime and let $m=p^2$, where $q$ is also a prime such that $p&lt;q$. This satisfies all your requirements.
Given a squarefree $a$ and $c&gt;b&gt;1$ such that $b$ is not a factor of $c,$ then we can define $m=ab^2$ need $n=abc.$ 

Then $n=abc&gt;ab^2=m,$ and $n^2=a^2b^2c^2$ is divisible by $m=ab^2$, and $\frac{n}{m}=\frac{abc}{ab^2}=\frac{c}{b}$ is not an integer since $c$ is not divisible by $b.$



On the other hand, given any two $m,n$ satisfying these conditions, we can find $a,b,c.$

If $m$ is square-free, then $m\mid n^2$ implies $m\mid n.$ (Why?) So $m$ is not square-free.

This means that we can write (uniquely) $m=ab^2$ where $a$ is square-free and $b&gt;1.$ 

Now, since $ab^2\mid n^2,$ with $a$ square-free, then then $ab\mid n.$ (Why?) 

Writing $n=abc$ for some $c,$ then we then see $n&gt;m$ means $c&gt;b.$ We also see that that $m=ab^2$ is a factor of $n=abc$ if and only if $b$ is a factor of $c.$ 

So this construction gives all $m,n$ uniquely: Start with $a$ square-free, and $c&gt;b&gt;1$  with $c$ not a multiple of $b.$ Then take $m=ab^2,n=abc.$



Simple cases: When $n=1$ and $b&gt;1$ we can choose $c=b+1$ then we get $m=b^2$ and $n=b(b+1).$