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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?

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    $\begingroup$ 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. $\endgroup$ Sep 10, 2021 at 11:01
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    $\begingroup$ Does the trivial solution $(x, y, z) = (0, 0, 0)$ not count for $2x^2+3y^2=z^2$? $\endgroup$ Sep 10, 2021 at 11:03
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    $\begingroup$ @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? $\endgroup$ Sep 10, 2021 at 11:08
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    $\begingroup$ artofproblemsolving.com/community/… $\endgroup$
    – individ
    Sep 12, 2021 at 14:16
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    $\begingroup$ @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. $\endgroup$ Sep 12, 2021 at 14:37

2 Answers 2

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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, $p<z$, which means $(r,q,p)$ is a smaller solution, which contradicts the first statement. Therefore, there are no solutions for $(x,y,z)$.

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  • $\begingroup$ $2x^2 + 3y^2 = z^2\quad$ has no non-zero real solutions $\endgroup$
    – poetasis
    Sep 10, 2021 at 15:12
  • $\begingroup$ @poetasis That is obviously not true. $\endgroup$ Sep 11, 2021 at 8:37
  • $\begingroup$ @ Servaes You are right. I ignored $(0,0,0)# $\endgroup$
    – poetasis
    Sep 11, 2021 at 15:20
  • $\begingroup$ @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) $\endgroup$
    – MafPrivate
    Sep 12, 2021 at 7:03
  • $\begingroup$ A mod $3$ argument shows $2x^2+3y^2=z^2$ implies $3\mid\gcd(x,y,z)$, which gives the same contradiction. $\endgroup$ Sep 18, 2021 at 11:49
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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 $(a,b,c,d)=(14, 1, 13, 2): p^2-13q^2 = 1.$
$(x,y,z)=(1, 2194919, 608761),(1, 3698003921039, 1025641750321)$...

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