On the existence of solutions to two integer quadratic equations of the same "parity"

Question

Let $a,b,c,d\in\mathbb{Z}$ such that $b\mid a^2+1$, $d\mid c^2+1$, $a\equiv c\,(2), b\equiv d\,(2)$.

I have the equations $$b^2 x^2-2abxy+y^2(a^2+1)=b,\quad d^2 x^2-2cdxy+y^2(c^2+1)=d$$ and I am guaranteed that there exist integer solutions $(x,y)$ to the first one and $(x', y')$ to the second one (Not necessarily the same), but I would like to know if always exist integer solutions $(x_1,y_1)$ of the first and $(x_2,y_2)$ of the second with $x_1\equiv x_2\,(2)$ and $y_1\equiv y_2\,(2)$

Analyzing cases I get to:

$a$ even, $b$ odd then \begin{cases}y \,\text{odd} \Rightarrow x\, \text{even}, \\y\,\text{even} \Rightarrow x \,\text{odd}\end{cases}

So in this case it works.

$a$ odd, $b$ odd then $x$ must be odd but I don't know nothing for $y$.

$a$ even, $b$ even cannot happen

$a$ odd, $b$ even gives me no information.

Thanks in advance!

Answer 1

Following sketch is based on idea that each of the equations must have two solutions that differ in parity, so if you want two equations that also have different parity, this results in four different parity combinations, yet in this case only three parity combinations are possible for any solution...

First let $P=b,Q=-2a,R=\frac{a^2+1}{b}$, after dividing by $b$ we have a diophantine equation $$ Px^2+Qxy+Ry^2=1\tag{*}. $$

This is a quadratic diophantine equation, you can search for generic solutions (there are some on this site too). In our case we have discriminant $D=-4$, so we can use for example Lagrange’s Algorithm Revisited: Solving $at^2 + btu + cu^2 = n$ in the Case of Negative Discriminant, see especially sections 2 and 4. To summarize, the equation has following solutions by cases:

• If $P=1$, $(x,y)=\pm(1,0),\pm(-N,1)$, notice the two solutions have different parity, specifically in second component
• If $P=2$, $(x,y)=\pm(\frac{-N+1}{2},1), \pm (\frac{-(N+1)}{2},1)$ for an odd integer $N$, again we can see the parity differs in first component
• Otherwise solutions are given by two consecutive convergents in continued fraction of $-\frac{Q}{2P}$ in a form $\pm (x,y)$ and $\pm (Nx+Ry,Px+Ny)$. Now to see that these two will differ in parity is a bit tricky, but can be done by inspecting three cases of parity of first solution $(x,y)$ (see below why only three...), then combine it with equation $(*)$ to show that $(Nx+Ry,Px+Ny)$ will have different parity.

We see that equation $(*)$ has by itself always at least two solutions with different parity.

Notice also that $(*)$ implies we cannot have both $x$ and $y$ even, so there are only three possibilities modulo $2$: $(1,0),(0,1)$ and $(1,1)$.

Finally, for the sake of contradiction assume we have a pair of equations with solutions that do not have same parity. That gives at least two distinct parity combinations from each of them, so at least four different parity possibilities together. That is clearly not possible as we have shown there can be at most three possibilities...