# Does the diophantine equation $ax^2+by^2=cz^2+d$ always have solutions?

Let $$a,b,c,d,x,y,$$ and $$z \in \mathbb{N}$$ where $$a,b,c,$$ and $$d$$ are constants but d is allowed to be zero .

$$ax^2+by^2=cz^2+d$$

First example :

when $$a=b=c=1$$ and $$d=0$$ we have the equation : $$x^2+y^2=z^2$$ which I know its general solution .

Second example:

$$a=1 ,b=4,c=1$$ and $$d=0$$ we have the equation : $$x^2+4y^2=z^2$$ . it has solutions.

one solution of it is: $$x=3,y=2,$$ and $$z=5$$

Third example:

$$a=2 ,b=3,c=1$$ and $$d=0$$ we have the equation : $$2x^2+3y^2=z^2$$ .

this example I tried to find solutions among small numbers but I didn't find any solution.

So does $$2x^2+3y^2=z^2$$ have solutions but I didn't find any ?

or is there proof that $$2x^2+3y^2=z^2$$ doesn't have any solution?

• gcd-type constraints will come in e.g. $4x^2+4y^z = 4z^2+1$ would have no solution, but otherwise I can't see anything stopping this. Sep 10, 2021 at 11:01
• Does the trivial solution $(x, y, z) = (0, 0, 0)$ not count for $2x^2+3y^2=z^2$? Sep 10, 2021 at 11:03
• @Mahmoudalbahar But you said "Let $a,b,c,d,x,y,$ and $z \in \mathbb{N}$" and later on had " $d=0$" so presumably your $\mathbb{N}$ includes $0$ or was that a mistake? Sep 10, 2021 at 11:08
• artofproblemsolving.com/community/… Sep 12, 2021 at 14:16
• @individ Thank you very much and your blog is really great because it contains many things about diophantine equations and number theory... but please I want you when it is convenient to you to make example here on mathstack using the method in your blog. Sep 12, 2021 at 14:37

Well $$2x^2+3y^2=z^2$$ actually has no solution for positive integers $$x,y,z$$.

Assume the smallest solution of $$(x,y,z)$$ (with minimal $$z$$) is $$(r,s,t)$$. Consider $$\mod 8$$, as $$2x^2+ 3y^2\equiv 0,2,3,4,5,6 \pmod{8}$$ and $$z^2 \equiv 0,1,4 \pmod{8}$$, we can see $$z^2 \equiv 0,4 \pmod{8}$$, implies that $$z$$ is even. Let $$z=2p$$, then $$2x^2+3y^2=4p^2$$, which also implies $$y$$ to be even. Let $$y=2q$$, then $$2x^2+12q^2=4p^2$$, which becomes $$x^2+6q^2=2p^2$$, implying $$x$$ is also even. Let $$x=2r$$, then the equation becomes $$4r^2+6q^2=2p^2$$, which is $$2r^2+3q^2=p^2$$. Therefore, $$(r,q,p)$$ is also a solution. However, $$p, which means $$(r,q,p)$$ is a smaller solution, which contradicts the first statement. Therefore, there are no solutions for $$(x,y,z)$$.

• $2x^2 + 3y^2 = z^2\quad$ has no non-zero real solutions Sep 10, 2021 at 15:12
• @poetasis That is obviously not true. Sep 11, 2021 at 8:37
• @ Servaes You are right. I ignored $(0,0,0)# Sep 11, 2021 at 15:20 • @poetasis There are infinitely many real solutions for the equation for obvious reasons. What I have proven is just nonzero integer solutions (which can be applied to nonzero rational solutions) Sep 12, 2021 at 7:03 • A mod$3$argument shows$2x^2+3y^2=z^2$implies$3\mid\gcd(x,y,z)\$, which gives the same contradiction. Sep 18, 2021 at 11:49

This is the partial solution.

$$ax^2+by^2-cz^2=d\tag{1}$$
Substitute $$x=1, y=pt+1, z=qt+1$$ to above equation.
$$p,q$$ are arbitrary.

If $$a+b=c+d$$ then we get $$t = \large\frac{-2(bp-cq)}{-cq^2+bp^2}.$$
If $$bp^2-cq^2=\pm1$$, we can get the integral solution $$(x,y,z).$$
Thus, equation $$bp^2-cq^2=\pm1$$ is reduced to Pell's equation $$X^2-bcY^2=\pm b\tag{2}.$$
If equation $$(2)$$ has a solution , then equation $$(1)$$ has infinitely many positive integer solutions.

In particular, let $$b=1$$ then if $$c$$ is not a perfect square, Pell's equation $$X^2-cY^2=1$$ always has infinitely many distinct integer solutions.

Example for $$(a,b,c,d)=(4,1,2,3): p^2-2q^2 = 1.$$
$$(x,y,z)=(1, 7, 5),(1, 239, 169),(1, 8119, 5741),(1, 275807, 195025)$$...

Example for $$(a,b,c,d)=(14, 1, 13, 2): p^2-13q^2 = 1.$$
$$(x,y,z)=(1, 2194919, 608761),(1, 3698003921039, 1025641750321)$$...