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

##### 1 Answer

This has integer solutions:

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

#### Explanation:

Look for integer solutions

Given:

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

Add

#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

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

So the only possible value of

Then

So

**Rational solutions**

Consider the sequence

#{ (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

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

Discarding the initial

#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