# How do you solve 1/a^2-1/b^2 = 3/4 ?

Sep 15, 2016

This has integer solutions:

$a = \pm 1$, $b = \pm 2$

#### Explanation:

Look for integer solutions $a , b$

Given:

$\frac{1}{a} ^ 2 - \frac{1}{b} ^ 2 = \frac{3}{4}$

Add $\frac{1}{b} ^ 2$ to both sides to get:

$\frac{1}{a} ^ 2 = \frac{3}{4} + \frac{1}{b} ^ 2 = \frac{3 {b}^{2} + 4}{4 {b}^{2}}$

Take the reciprocal of both sides to get:

${a}^{2} = \frac{4 {b}^{2}}{3 {b}^{2} + 4}$

So if $a$ is an integer, then $3 {b}^{2} + 4$ is a divisor of $4 {b}^{2}$, but for $b \ge 2$ we have:

$4 {b}^{2} \ge 3 {b}^{2} + 4 > 2 {b}^{2}$

So the only possible value of ${a}^{2}$ is $1$.

Then $\frac{1}{1} - \frac{1}{b} ^ 2 = \frac{3}{4}$, hence $\frac{1}{b} ^ 2 = \frac{1}{4}$, hence ${b}^{2} = 4$

So $a = \pm 1$ and $b = \pm 2$

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Rational solutions

Consider the sequence ${b}_{0} , {b}_{1} , {b}_{2} , \ldots$ defined as follows:

$\left\{\begin{matrix}{b}_{0} = 0 \\ {b}_{1} = 2 \\ {b}_{n + 1} = 4 {b}_{n} - {b}_{n - 1}\end{matrix}\right.$

The first few terms are:

$0 , 2 , 8 , 30 , 112 , 418 , \ldots$

Then $3 {b}_{n}^{2} + 4$ is a square number.

(See https://socratic.org/s/axXmkYA6 for a proof)

Discarding the initial $0$, which leads to a zero denominator in our original equation, we have rational solutions $\left({a}_{1} , {b}_{1}\right)$, $\left({a}_{2} , {b}_{2}\right)$,... where:

${a}_{n} = \sqrt{\frac{4 {b}_{n}^{2}}{3 {b}_{n}^{2} + 4}} = \frac{2 \left\mid {b}_{n} \right\mid}{\sqrt{3 {b}_{n}^{2} + 4}}$

There are other solutions for rational, non-integer values of $b$, but these are all the positive solutions for integral $b$.