Question

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

Answer 1

This has integer solutions:

`a = +-1`, `b = +-2`

Explanation:

Look for integer solutions `a, b`

Given:

`1/a^2-1/b^2 = 3/4`

Add `1/b^2` to both sides to get:

`1/a^2 = 3/4+1/b^2 = (3b^2+4)/(4b^2)`

Take the reciprocal of both sides to get:

`a^2 = (4b^2)/(3b^2+4)`

So if `a` is an integer, then `3b^2+4` is a divisor of `4b^2`, but for `b >= 2` we have:

`4b^2 >= 3b^2+4 > 2b^2`

So the only possible value of `a^2` is `1`.

Then `1/1-1/b^2 = 3/4`, hence `1/b^2 = 1/4`, hence `b^2=4`

So `a=+-1` and `b=+-2`

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Rational solutions

Consider the sequence `b_0, b_1, b_2,...` defined as follows:

`{ (b_0 = 0), (b_1 = 2), (b_(n+1) = 4b_n - b_(n-1)) :}`

The first few terms are:

`0, 2, 8, 30, 112, 418,...`

Then `3b_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 `(a_1, b_1)`, `(a_2, b_2)`,... where:

`a_n = sqrt((4b_n^2)/(3b_n^2+4)) = (2abs(b_n))/sqrt(3b_n^2+4)`

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