Trying to prove that there are no p and q such that $|\sqrt5 - p/q| < 1/(7q^2)$. Like the title says, I'm having trouble proving that there are no integers p and q such that
$|\sqrt5 - p/q| < 1/(7q^2)$. I was given the hint that $|(q\sqrt5 - p)(q\sqrt5 + p)| \geq 1$, but I don't quite know how that helps...
Thanks!
 A: Since $\sqrt{5}$ is not a rational number we know that $5 n^2-m^2$ is a nonzero integer for every integers $n$, $m$ with $q\ne0$ thus
$$\forall\,(n,m)\in\Bbb{Z}^2\setminus\{(0,0)\},\quad 1\leq|5n^2-m^2|\tag{1}$$ 
Now, suppose that there is $(p,q)$ with $q\ne0$ such that
$$\left\vert\sqrt{5}-\frac{p}{q}\right\vert\leq\frac{1}{7q^2}$$
It follows that
$$\left\vert q\sqrt{5}-p\right\vert\leq\frac{1}{7|q|}\quad \hbox{and} \quad 
\left\vert q\sqrt{5}+p\right\vert\leq2|q|\sqrt{5}+\frac{1}{7|q|}$$
Thus
$$\left\vert 5q^2 -p^2\right\vert\leq\frac{1}{7|q|}\left(2|q|\sqrt{5}+\frac{1}{7|q|}\right)
= \frac{2\sqrt{5}}{7}+\frac{1}{49q^2} \leq
 \frac{2\sqrt{5}}{7}+\frac{1}{49}<\frac{6}{7}+\frac{1}{7}=1
$$
This contradicts $(1)$, and proves that no such $(p,q)$ exists.$\qquad\square$
A: I think you are my classmate, since we had submitted our homework a few hours ago, I would like to share my solution. Because we are studying continued fractions now, so my solution is based on it.
First, we know if we can find some $(p, q)$ pair, and $q \not= 0$, by Theorem 12.18 or Corollary 12.18.1 in the textbook(Elementary Number Theory by Kenneth H. Rosen 6ed), then the most possible solutions are the convergents $C_k = \frac{p_k}{q_k}, k = 0, 1, 2, ...$ of $\sqrt{5}$. Because they are the closest rational approximation if $q$ is given. But we can show no one in those convergents satisfy this inequality. 
We can easily know $\sqrt{5} = [2;\bar{4}]$ and $2 + \sqrt{5} = [\bar{4}]$.
When $k = 0$ and $k = 1$ we can just do some calculations:
$k = 0$, $|\sqrt{5} - \frac{2}{1}| = \sqrt{5} - 2 > \frac{1}{7}$, since $5 > \frac{225}{49}$.
$k = 1$, $|\sqrt{5} - \frac{9}{4}| = \frac{9}{4} - \sqrt{5} > \frac{1}{112}$, since $\frac{251}{112} > \sqrt{5}$.
For $k > 1$, We know $\sqrt{5} = \frac{\alpha_{k+1}p_k + p_{k-1}}{\alpha_{k+1}q_k + q_{k-1}}$, so 
$$\begin{aligned} |\sqrt{5} - \frac{p_k}{q_k}| &= |\frac{\alpha_{k+1}p_k + p_{k-1}}{\alpha_{k+1}q_k + q_{k-1}} - \frac{p_k}{q_k}| \\
&= |\frac{\alpha_{k+1}p_kq_k + p_{k-1}q_k - \alpha_{k+1}p_kq_k - p_kq_{k-1}}{\alpha_{k+1}q_k^2 + q_{k-1}q_k}| \\ &= |\frac{-(p_kq_{k-1} - p_{k-1}q_k)}{\alpha_{k+1}q_k^2 + q_{k-1}q_k}| \\ 
&= \frac{1}{\alpha_{k+1}q_k^2 + q_{k-1}q_k} \\
&> \frac{1}{\alpha_{k+1}q_k^2 + q_k^2}\end{aligned}$$
We know $\alpha_{k+1} = [\bar{4}] = 2 + \sqrt{5}$ so
$$|\sqrt{5} - \frac{p_k}{q_k}| > \frac{1}{(3 + \sqrt{5})q_k^2} > \frac{1}{7q_k^2}$$
Which completes the proof. $\blacksquare$
A: An alternative, we have $|q \sqrt{5} - p|< \tfrac{1}{7q}$ so if $ (q \sqrt{5}-p)(q \sqrt{5}+p) \geq  1 $ we can divide both sides
\begin{eqnarray}
\\
|q \sqrt{5} + p| & \geq &  7q \\
| \sqrt{5} + \tfrac{p}{q}| &\geq & 7
\end{eqnarray}
This is impossible since $| \sqrt{5} + \tfrac{p}{q}| \leq \sqrt{5} + \tfrac{p}{q} \leq \sqrt{5}+1 < 4$.
