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Find the number of pairs $(m, n)$ of positive integers such that $\frac{ m}{n+1} < \sqrt{2} < \frac{m+1}{n}$ Constraint: $m$ and $n$ are both less than or equal to 1000

I toiled over this problem for a bit. I tested the first few positive integers of m and found the corresponding values of $n$, but no real pattern seemed to emerge. Clearly $m \ge n$ . I believe that for each value of $m$, there were either two or one value(s) for $n$ in no clear order, which is the part that is messing me up. This is meant to be done by hand so I have not used any software on it.

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    The answer is exactly $999+\lfloor{1001/\sqrt{2}\rfloor}$, and the pattern is followed as described in my answer below - this works in general.2012-12-26

2 Answers 2

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You can rewrite this as: $\frac{m}{\sqrt{2}}-1

Since $\sqrt{2}$ is irrational, we know that $\frac{m}{\sqrt{2}}-1$ is not an integer, so for an integer $n>\frac{m}{\sqrt{2}}-1$ iff $n\geq \left\lceil \frac{m}{\sqrt{2}}-1\right\rceil =\left\lfloor \frac{m}{\sqrt{2}}\right\rfloor$.

Similarly, $n<\frac{m+1}{\sqrt{2}}$ iff $n\leq \left\lfloor \frac{m+1}{\sqrt{2}}\right\rfloor$. So altogether, we are seeking $m,n$ such that $\left\lfloor\frac{m}{\sqrt 2}\right\rfloor\leq n\leq \left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor$

For a particular $m$, then, the number of possible $n$ is: $\left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor-\left\lfloor\frac{m}{\sqrt 2}\right\rfloor +1$. Summing over all $m$, the result is $-1+\sum_{m=1}^{1000}\left(\left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor-\left\lfloor\frac{m}{\sqrt 2}\right\rfloor +1\right)$

(The $-1$ is because we don't want to count $n=0$ when $m=1$.)

But this is just $999$ plus a telescoping sum, and we see that the result is:

$999+\left\lfloor\frac{1001}{\sqrt{2}}\right\rfloor$

Actually, even more specifically, it is:

$1000-\lfloor\sqrt{2}\rfloor + \left\lfloor\frac{1000+1}{\sqrt{2}}\right\rfloor$

This will work for any irrational number $\alpha>1$ and any upper bound, $M>\alpha$, yielding a total: $M-\lfloor\alpha\rfloor + \left\lfloor\frac{M+1}{\alpha}\right\rfloor$

which counts the pairs $(m,n)$ with $1\leq m,n\leq M$ and $\frac{m}{n+1}<\alpha<\frac{m+1}n$

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    So if $n$ is an integer and $x$ is not, then n>x-1 if and only if $n\geq \lfloor x\rfloor$. Thus, this technique definitely requires that $\sqrt{2}$ is irrational.2012-12-26
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I don't understand. Why such complexity?

Inequality: $\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n}$

This is equivalent to the inequality :

$n\sqrt{2}-1

It is easy to see that one is " $n$ " - corresponds to one value " $m$ ".

Or let's different. $\frac{m}{\sqrt{2}}-1

Set the number " $m$ " - and integer " $2n$ " - and will be the number of all possible pairs.

$1+\sqrt{2}=2,4142$

So the number of all possible pairs given " $m$ ". 't care. " $ 2n $ " .