# How to Find Integer Solution for $\frac {x^2-y^2}{a^2-b^2}$

Given $$\frac {x^2-y^2}{a^2-b^2}$$ is an integer,

For which non-zero $$x, y, a, b\in\mathbb Z$$, the fraction $$\frac {x^2-y^2}{a^2-b^2}$$ is an integer?

I can start with $$x^2-y^2 = (x-y)(x+y)$$ but it does not lead anywhere.

• @ViktorGlombik Thats a good idea, I might try this time, no, it is not a exercise, I saw various time such expressions, and curious to know how such cases are solved for integer solutions Jan 29, 2020 at 18:00
• @ViktorGlombik am looking for a closed form formula, brute force or other type of programming is not what I am interested. Jan 29, 2020 at 18:04
• Notice that for since you are taking squares it suffices to consider $x,y,a,b \ge 0$. Some solutions are given by $\{ (a k,0,k,0): a, k \in \mathbb Z\}$. Jan 29, 2020 at 19:05
• Also, there are many solutions. For $x,y,a,b \in \{0, \ldots, 100\}$ and $x > y$ and $a > b$ there are $138790$ solutions. Jan 29, 2020 at 20:33

Above equation shown below is equivalent to:

$$x^2-y^2=w(a^2-b^2)$$ ----$$(1)$$

where '$$w$$' is an integer

Equation (1) has parametric solution given below:

$$x=(k+3)(k^2-1)$$

$$y=4k(k+1)$$

$$a=(k^2+4k-1)$$

$$b=2(k-1)$$

$$w=(k+1)(k-3)$$

For, k=4 we get, $$(x,y,a,b)=(105,80,31,6)$$

and, $$w=5$$

• Upvoted ur answer but not confirm it is correct, how u obtain? what is the method? Jan 30, 2020 at 0:16

Here is one description of the solution set that may be unsatisfying.

Fix any $$x,y$$. Then $$x^2-y^2$$ has some finite list of factors. For each factor in that list, either $$a^2-b^2=(\text{that factor})$$ has integer solutions for $$a$$ and $$b$$ or it does not. There is a solution exactly when that factor is not congruent to $$2$$ mod $$4$$. When there is a solution, there may be multiple solutions, but not infinitely many.

So the solution set can be described as:

• Let $$x$$ be any integer.
• Let $$y$$ be any integer.
• For each factor $$d$$ of $$x^2-y^2$$ where $$d\not\equiv 2$$ mod $$4$$, take all solutions to $$a^2-b^2=d$$.