Find all integer solutions $a, b, c$ to $ab - bc + ca = 1$

Question

(1) Are there infinitely many integer solutions $(a, b, c)$ to $ab - bc + ca = 1$?

(2) If so, is there a good method for finding any?

(3) Please give other examples if possible.

For example; $a = 2, b = 3, c = 5$ is a solution to $ab - bc + ca = 1$ because $2\cdot 3 - 3\cdot 5 + 2\cdot 5 = 1$.

Also $\gcd(a, b, c) = 1$ should be true.

Answer 1

Here, I am looking for positive integer solutions. We have $$\frac{c^2-1}{b+c}=c-a\,.$$ So, for a fixed integer $c>1$, take a divisor $d>c$ of $c^2-1$ (which exists since $c^2-1>c$). Then, $$b=d-c\text{ and }a=c-\frac{c^2-1}{d}$$ are positive integers. This method gives all solutions $(a,b,c)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}\times\mathbb{Z}_{>1}$. Furthermore, for each $c>1$, there are exactly $$\frac{1}{2}\,\tau\left(c^2-1\right)$$ pairs $(a,b)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}$ which satisfy the required condition. (Here, $\tau$ is the divisor-counting function.)

In particular, $(a,b,c)=(1,1,c)$ works for every $c$ (this corresponds to $d=c+1$). Donald Splutterwit's family of solutions corresponds to $c=2n-1$ with $d=4n$ (thus, $b=d-c=2n+1$ and $a=c-(n-1)=n$). Another infinite family is when $d=c^2-1$, so that $$(a,b,c)=\left(c-1,c^2-c-1,c\right)\,.$$

Answer 2

$a = b = 1 \longrightarrow 1 - c + c = 1 \longrightarrow 1=1$

Answer 3

$a=n, b=2n+1 , c=2n-1$ gives an infinite family of solutions.

Answer 4

Your form is $SL_3 \mathbb Z$ equivalent to $y^2 - zx.$ All solutions to $y^2 - zx = 1$ are given by $y^2 - 1 = zx.$ Meaning $y$ can be anything and then we demand $xz = y^2 - 1$

Yep. In one direction, $$ a = y, \; \; b = x+y, \; \; c = y+z. $$ Backwards $$ x = b-a, \; \; y = a, \; \; z = c-a. $$