Problem: Given that $m, n$ are positive integers such that $\sqrt{7} -\frac{m}{n} > 0$. Then show that $\sqrt{7}-\frac{m}{n} > \frac{1}{mn}$.

I have failed to do this fascinating problem.

My efforts: I tried to approach by contradiction. Assume that we have $\sqrt{7} -\frac{m}{n} > 0$ and $\sqrt{7} -\frac{m}{n} > \frac{1}{mn}$ both hold simultaneously.

Now, $\sqrt{7} -\frac{m}{n} > \frac{1}{mn}$ implies $(m-\frac{n\sqrt{7}+\sqrt{7n^2-4}}{2})(m-\frac{n\sqrt{7}-\sqrt{7n^2-4}}{2}) >0$.

Since, $m \geq 1$ and $\frac{n\sqrt{7}-\sqrt{7n^2-4}}{2} < 1 $.

We end up with, $\frac{n\sqrt{7}+\sqrt{7n^2-4}}{2} < m < n\sqrt{7}$.

So, now I need to show that the last inequality cannot hold for positive integers $m, n$. But I am unable to do that.

So, someone please help me. Please..


marked as duplicate by Gabriel Romon, user147263, John B, Shailesh, zz20s Apr 12 '16 at 0:09

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A more direct approach.

We write:

$$1\leq 7n^2-m^2 = (n\sqrt{7}+m)n \left(\sqrt{7}-\frac{m}{n}\right)$$

If $\sqrt{7}-\frac{m}{n}\lt \frac{1}{mn}$ (equality is not possible) then $n\sqrt{7}< m+\frac{1}{m}$ so $m+n\sqrt{7}\lt 2m+\frac{1}{m}$ and $7n^2-m^2\lt (2m+\frac{1}{m})n\frac{1}{mn}=2+\frac{1}{m^2}$.

So $7n^2-m^2=1\text{ or }2$. But $-1$ and $-2$ are not squares modulo $7$.

I doubt you will be able to get this without looking at congruences, because, for example, $m^2-Dn^2=-1$, then $$0<\sqrt{D}-\frac{m}{n}=\frac{1}{n(m+n\sqrt{D})}<\frac{1}{mn}$$ for these values, but when, say prime $p\equiv 7\pmod{8}$ then you never have $0<\sqrt{p}-\frac{m}{n}<\frac{1}{mn}$. So the general result is at least partly about these equations.


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