Is it easier to find $a^2-8c=b^2$ than $a^2-c=b^2$

Question

I found a way to factor numbers if I find:

$$a^2-8c=b^2$$

• Where $c$ is the number I want to factor

Is it easier than searching for the next equation?

$$a^2-c=b^2$$

Answer 1

If you have $a^2-c^2 = b^2$, then you can use the known result about solutions to $u^2+v^2 = w^2$: there exists $m$ and $n$ such that $u = m^2-n^2$, $v = 2mn$, and $w = m^2+n^2$.

For $a^2-8c^2 = b^2$, I am not aware of a similar result, though one might exist.

In both cases, you can look at the factorization of $a^2-b^2 =(a-b)(a+b) $. The presence of the $8$ would force both $a+b$ and $a-b$ to be even (since of one of them is even the other is also, and similarly for odd), and would also force the power of two that divides one of them to be even and the other to be odd.

In other words, once you have $c^2 = a^2-b^2$ or $8c^2 = a^2-b^2$, your factoring job is essentially over, unless you want the complete factorization of very large numbers.