# find all rational numbers p/q such that $|p/q-17/5|< 1/q^2$

This is a question in my assingment. I needto find all rational numbers p/q such that $|p/q-17/5|< 1/q^2$. Any ideas ? Thanks for any help!

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Multiply both sides by $5q^2$, and ask yourself how it can happen that the product of two integers is less than 5 in absolute value.

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You need to convert this question into more suitable form: Given condition is equivalent to $\frac{|5p - 17q|}{|5q|} < \frac{1}{|q|^2}$ where p and q are integers. You can always multiply or divide both sides of inequality by any positive number.

Finally you would get, $|5p - 17q||q| < 5$, assuming we are looking for non-trivial solution (p/q = 17/5 is a trivial solution), this can only be for $q \in [-4,4]$. If q = +/-4, $|5p =17q|$ could only be +/-1. similarly for others and you would get finite many valid (p,q) pairs.

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$\left |\cfrac pq - \cfrac {17}5 \right | < \cfrac 1{q^2} \quad \Rightarrow -\cfrac 1{q^2} < \cfrac pq - \cfrac {17}5 < \cfrac 1{q^2}$. Now you just have to find all set of integers $(p, \space q)$ that satisfy these inequalities

$$\begin{array}{l l} 1.\quad -\cfrac 1{q^2} < \cfrac pq - \cfrac {17}5\\ 2. \quad \cfrac pq - \cfrac {17}5 < \cfrac 1{q^2}\\ \end{array}$$

If you multiply both inequality by $5q^2$ you get this $$-5<q (5 p-17 q)<5$$ This is easy because $5$ is a prime number and $q$ and $5 p-17 q$ are both integers, so you have to consider that $-5 \le q \le 5$ and then you can then study the inequality when $$q \in \{-1,\space 1,\space -2,\space 2,\space -3 ,\space 3,\space -4 , \space 4\, \space -5 , \space 5\}$$

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What about (p,q)=(51,15)? –  Did Oct 28 '12 at 14:02
that's a multiple of $\cfrac {17}{5}$ and $\cfrac {17}{5}$ is an obvious solution. –  user31280 Oct 28 '12 at 14:10
Sure. But this shows there is a case missing in your answer. –  Did Oct 28 '12 at 14:11
my interval should have been $-5\le q \le 5$. Thanks! –  user31280 Oct 28 '12 at 14:13
No. (Note that if (p,q)=(51,15), then q is NOT between -5 and 5.) –  Did Oct 28 '12 at 14:15