# Finding solutions if existed in this number theory - prime numbers problem.

I was working on a problem involving prime number and I reached a point where I could not solve an equation, and the equation is. Given that $$a,b$$ and $$c$$ are positive integers which satisfies that both $$a$$ and $$b$$ are less than $$c$$. And satisfy the following equation $$ac=b^2 +1$$ Find solutions if theres any, if there exists infintly many solutions find $$a$$ and $$ab$$ as functions in tearms of $$c$$.

(The original problem was if given a number $$s$$, find the biggest prime divisor for ($$s^2 +1$$). And after some conclusions I came up with the above equation and didnt know how to find a solution, so if the first problem cannot be answered I hope that the second one can be).

• Welcome to Math StackExchange. The questions you ask are more or less about factoring S^2+1. This is an extremely difficult question. For instance when I was young, I was told it is unknown whether there exists infinitely many values of S such that S^2+1 is prime. I think the question is still open today. Commented Dec 22, 2019 at 22:08
• $AC = B^2+1$ has solutions whenever $B^2+1$ is composite. Examples are $B^2+1=10,26,50,65,82,122$ etc. As for $A,B$ in terms of $C$ than $B=\sqrt{AC-1}$ and $A = \frac {B^2+1}C$. I don't see why you'd assume that infinite number of solutions would mean solutions for every $C$..... Commented Dec 22, 2019 at 22:18
• @arnaud. This is the exact opposite. We are looking for examples where $S^2 + 1$ is composite and it is easy to know there are infinitely many of those. Commented Dec 22, 2019 at 22:19
• Thank you both for answering but let me tell you more about this problem, intially S had specific valuse S=111 and 421 and the question was was find the biggest prime divisor for 111^2 +1 and 421^2 +1 so i though why not generalize the idea and i did so, so the question became how can we relate between the prime divisors of $S^2+1$ and S it self, i changed P the prime divisor of $S^2+1$ to C and i made it into a number theory problem in hope of finding a relation between S and P (remember P is the prime divisor of $S^2+1$ ), in the end can we find such relation and how so? Commented Dec 22, 2019 at 22:33
• And if we can find a relation between P and S ,can we express P in terms of S or the other way around? Commented Dec 22, 2019 at 22:36

Changing your letters to become $$y^2 - z x = -1.$$ The automorphisms (with positive determinant) of the quadratic form $$y^2 - zx$$ are known, and parametrized by the modular group. So, given $$ps-qr = 1,$$ so $$\det \left( \begin{array}{cc} p & q \\ r & s \\ \end{array} \right) = 1$$ we construct $$W = \left( \begin{array}{ccc} p^2 & 2pq & q^2 \\ pr & ps+qr& qs \\ r^2 & 2rs & s^2 \\ \end{array} \right)$$ The Hessian matrix of the quadratic form is $$H = \left( \begin{array}{ccc} 0 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 0 \\ \end{array} \right)$$ and we have arranged $$W^T H W = H.$$ This means that, if $$\left( \begin{array}{c} x \\ y \\ z \\ \end{array} \right)$$ solves $$y^2 - zx = -1,$$ so does every $$\left( \begin{array}{ccc} p^2 x + 2pq y + q^2 z \\ pr x + (ps+qr) y + qs z \\ r^2 x +2rs y + s^2 z \\ \end{array} \right)$$ For example, if we take $$x=1, y=0, z=1$$ to begin, $$\left( \begin{array}{ccc} p^2 + q^2 \\ pr + qs \\ r^2 + s^2 \\ \end{array} \right)$$ is a solution vector whenever $$ps-qr = 1$$