If $ab^2+1 = c^2+d^2$ with $a$ squarefree, what [else] can be said about $a$?

Question

What is known about squarefree integers $a$ where there exist non-zero integers $b$, $c$, and $d$, with $\gcd(b,c)=\gcd(b,d)=1$, such that $$ab^2+1=c^2+d^2$$ ?

EDIT: As pointed out by individ, if an integer $g>1$ exists such that $g \mid \gcd(b,c)$, then $b=gb_1$ and $c=gc_1$ yields \begin{align} a(gb_1)^2 + 1 &= (gc_1)^2+d^2 \\ d^2 - (ab_1^2-c_1^2)g^2 = 1. \end{align} Now if $ab_1^2-c_1^2$ is not a square, there are infinite solutions. So to make the question less trivial, we may assume $\gcd(b,c)=\gcd(b,d)=1$.

Answer 1

Under the condition $\gcd(b,cd)=1$, one thing we can say for sure is that, if there is a solution to $ab^2+1=c^2+d^2$, then $a\equiv0$, $1$, $3$, $4$, or $7$ mod $8$. This is because $b$ can never be even, since $2\mid b$ implies $c$ and $d$ are both odd, which makes $c^2+d^2\equiv2$ mod $4$ whereas $ab^2+1\equiv1$ mod $4$. But if $b$ is odd, then $ab^2+1\equiv a+1$ mod $8$, whereas $c^2+d^2\equiv1$, $2$, $4$, $5$, or $0$ mod $8$.

Another way to say this is that there is definitely no solution if $a\equiv2$ mod $4$ or $5$ mod $8$. If I had to guess, I'd say there probably are solutions for all $a$ in the other congruence classes.

Remark: The condition $cd\not=0$ as well as $\gcd(b,cd)=1$ makes life a little more interesting than it would be if we allowed solutions of the form $(b,c,d)=(1,c,0)$. For example, for $a=8$, $(b,c,d)=(1,3,0)$ would be an easy solution, but if it's disallowed, the next solution (i.e., the smallest value of $b$ that works) is $8\cdot17^2+1=48^2+3^2$.

Answer 2

Generally speaking for the equation. $$x^2+y^2=qz^2+1$$ For all values of the coefficient $q$ There are solutions. It is easy to show.

Will make a replacement. $y=as$ ; $z=bs$ Then.

$$x^2-(qb^2-a^2)s^2=1$$

This equation Pell. And we can always choose such numbers $a,b$ . To the expression $(qb^2-a^2)$ It was not square.

Answer 3

I thought it might be worth setting up a community wiki to record solutions to $ab^2+1=c^2+d^2$ with $\gcd(b,cd)=1$ and $cd\not=0$, for positive values of $a$ congruent to $0$, $1$, $3$, $4$, and $7$ mod $8$ (which my other answer established as the only congruence classes for which solutions might exist), and with $b$ as small as possible. I hope others will extend the list and/or correct any mistakes I (or others) may have made. (It's easy enough to check if an entry is a correct solution, but it can be tricky to check that it uses the smallest possible $b$ -- e.g., why doesn't anything less than $b=17$ work for $a=8$?)

$$\begin{align} 1\cdot1^2+1 &=1^2+1^2\\ 3\cdot7^2+1 &=12^2+2^2\\ 4\cdot1^2+1 &=2^2+1^2\\ 7\cdot1^2+1 &=2^2+2^2\\ 8\cdot17^2+1 &=48^2+3^2\\ \\ 9\cdot1^2+1 &=3^2+1^2\\ 11\cdot11^2+1 &=36^2+6^2\\ 12\cdot1^2+1 &=3^2+2^2\\ 15\cdot13^2+1 &=50^2+6^2\\ 16\cdot1^2+1 &=4^2+1^2\\ \\ 17\cdot1^2+1 &=3^2+3^2\\ 19\cdot1^2+1 &=4^2+2^2\\ 20\cdot7^2+1 &=30^2+9^2\\ 23\cdot23^2+1 &=78^2+78^2\\ 24\cdot1^1+1 &=4^2+3^2\\ \\ 25\cdot1^2+1 &=5^2+1^2\\ 27\cdot19^2+1 &=98^2+12^2\\ 28\cdot1^2+1 &=5^2+2^2\\ 31\cdot1^2+1 &=4^2+4^2\\ 32\cdot13^2+1 &=72^2+15^2\\ \\ 33\cdot1^2+1 &=5^2+3^2 \end{align}$$