I would like to show that for positive integers $a>b,c$ all greater than 1 such that $c\nmid a$, there are no solutions to the following equation:
$$a^2+1=b^2(c^2+1)$$
As was pointed out in the comments, it might help to rearrange and factor:
$$(a-b)(a+b)=(bc+1)(bc-1)$$
Another interesting approach to the question, if we rewrite:
$$a^2-b^2c^2=b^2-1$$
Then we are trying to argue that $b^2-1$ cannot be squarefree (in particular, $c^2|b^2-1$)
No. On the other hand, it appears that $c | a.$
I am watching a longer computer run, one infinite family turns out to be $$ a = 4 c^3 + 3 c, \; \; \; b = 4 c^2 + 1$$ for any positive integer $c.$ You can check that $a^2 + 1 = b^2 (c^2 + 1).$
There are a few others, though, with especially small $c,$ which makes me think they might be derivable from the infinite sequence.
EEDDIITTTTT: that worked. I have a doubly infinite sequence, one infinite sequence for a given $c.$
$$ a_0 = c, \; \; \; b_0 = 1, $$ $$ a_1 = 4 c^3 + 3 c, \; \; \; b_1 = 4 c^2 + 1,$$ $$ a_2 = 16 c^5 + 20 c^3 + 5 c , \; \; \; b_2 = 16 c^4 + 12 c^2 + 1,$$ then $$ \color{magenta}{ a_{n+2} = (4 c^2 + 2) a_{n+1} - a_n, \; \; b_{n+2} = (4 c^2 + 2) b_{n+1} - b_n}. $$ I believe already that this doubly indexed family gives all solutions.
a b c
38 17 2
117 37 3
268 65 4
515 101 5
682 305 2
882 145 6
1393 197 7
2072 257 8
2943 325 9
4030 401 10
4443 1405 3
5357 485 11
6948 577 12
8827 677 13
11018 785 14
a b c
The conjecture is true. Fix $c$. The equation is equivalent to $$ a^2 - (c^2+1) b^2 = -1 $$ which is a "Pell equation", or in modern terminology the condition for $u = a + b \sqrt{c^2+1}$ to be a unit of norm $-1$ in ${\bf Z}[\sqrt{c^2+1}]$. By inspection we find a unit $\epsilon = c + \sqrt{c^2+1}$ of norm $-1$ (corresponding to the easy solution $(a,b)=(1,c)$). This unit is fundamental because the coefficient of $\sqrt{c^2+1}$ in $\epsilon$ is $1$, which is as small as a positive integer can get. Hence every unit has the form $u = \epsilon^n$ for some odd integer $n$, and the units with $b>0$ are those for which the exponent $n$ is positive.
In the binomial expansion of $\epsilon^n = (c + \sqrt{c^2+1})^n$ the terms that contribute to $a$ are ${n \choose 2m} c^{n-2m} (c^2+1)^m$ for $0 \leq m < n/2$. Each of those is clearly divisible by $c$, so $c\,|\,a$ as claimed.
This also confirms Will Jagy's belief that his "doubly infinite sequence" gives all solutions: his solutions indexed $0,1,2,3,\ldots$ correspond to $\epsilon, \epsilon^3, \epsilon^5, \epsilon^7$ etc.
Unclear.
This equation is equivalent to the following : $a^2-qb^2=a^2-(c^2+1)b^2=-1$
You can use formulas solutions binary quadratic forms. It can be seen that the solutions of this equation is always there.
Though it is necessary to bring the decisions some pretty simple solutions:
the equation: $aX^2+bXY+cY^2=f$
If the root of the whole: $\sqrt{\frac{f}{a+b+c}}$
Then use the solution of Pell's equation: $p^2-(b^2-4ac)s^2=1$
Solutions can be written:
$Y=((4a+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$
$X=(-(4c+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$
If a root: $\sqrt{fa}$ then the solutions are of the form:
$Y=4ps\sqrt{fa}$
$X=(-2bps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a}}$
Although it should be mentioned, and the equation: $aX^2-qY^2=f$
If the root of the whole: $\sqrt{\frac{f}{a-q}}$
Using equation Pell: $p^2-aqs^2=1$ solutions can be written:
$Y=(2aps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$
$X=(2qps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$
And for that decision have to find double formula.
$Y_2=Y+2as(qsY-pX)$
$X_2=X+2p(qsY-pX)$
So, take that formula and believe. Hopefully this will be enough.
