Csdrhrt

For example ,the diophantine equation
$$x^2+1=25y$$
we can solve this by finding particular solution $(x,y)=(7,2)$
and using this , we can get general solution.

My question is
"To solve $ax^2+by+c=0$ $(a, b, c in mathbb{Z})$ we must find particular solution or
not?"

Assume $b>0$. If $b<0$ make a change of variables $y'=-y$. Let $d=gcd(a,b)$, then $d$ must divide $c$ so we can reduce ${a,b,c}$ to ${a/d,b/d,c/d}$. Hence we assume that $gcd(a,b)=1$.

Next, suppose $d = gcd(b,c)$. Then since $gcd(a,b)=1$, $d$ must divide $x^2$. Let $d=u^2v$ so that $v$ is squarefree, then $x$ must be of the form $x=uvw$ for some integer $w$, so we can replace the equation with
$$
begin{align}
ax^2 + by + c &= 0\
afrac{u^2v^2w^2}{d} + frac{b}{d} y + frac{c}{d} &= 0\
(av) w^2 + (b/d) y + (c/d) &= 0
end{align}
$$
So we replace ${a,b,c}$ by ${av,b/d,c/d}$.

After these two replacements, $gcd(b,c)=1$. If $d=gcd(a,b)neq 1$, then $d$ must divide $c$. This contradicts $gcd(b,c)=1$, so there are no solutions and we are done.

Hence for the remaining case we may assume that $gcd(a,b)=gcd(b,c)=1$. Now the problem can be reduced to solving
$$
ax^2 + c equiv 0 pmod b implies x^2 equiv -ca^{-1} pmod b
$$
i.e. a solutions exist iff $-ca^{-1}$ is a quadratic residue $pmod b$.

If $x$ is a solution so is $x+kb$ for any $kinmathbb Z$, so the distinct infinite classes can be classified by distinct solutions $0leq x < b$ (which can be none). This is a finite search which in some sense solves the problem.

For the given example it would be $x=7,18$ for the base case $0leq x < 25$, so the two infinite classes are $x=7+25k$ and $x=18+25k$ for $kinmathbb Z$.

To expand on this more, let the unique prime factorization of $b$ be
$$
b = prod_{i=1}^kp_i^{e_i}
$$

Then the equation can be split into
$$
begin{align}
x^2 &equiv -ca^{-1} pmod{p_i^{e_i}},quad 1leq i leq k
end{align}
$$
By solving each equation, the base solutions can be combined/obtained via Chinese Remainder Theorem. Also note that $a,cnot equiv 0 pmod{p_i}$ since $gcd(a,b)=gcd(b,c)=1$ by construction, so $xnotequiv 0 pmod{p_i}$.

Edit 1: The discussion below works for odd primes but gets tricky for even prime powers. If $2^t$ divides $b$ then the equation
$$
x^2 equiv -ca^{-1} pmod{2^t}
$$
needs to be considered separately.

For each odd prime, as $-ca^{-1} notequiv 0 pmod {p_i}$, then a solution to
$$
x^2 equiv -ca^{-1} pmod {p_i}
$$
depends on whether $-ca^{-1}$ is a quadratic residue $pmod{p_i}$. This will give either zero or two solutions $pmod{p_i}$. If any one of the equations has no solutions, then there must be zero solutions for the original equation.

Otherwise, each equation $pmod{p_i}$ has two solutions. Then since $xnotequiv 0 pmod{p_i}$, by Hensel's lifting it can be shown that there are also exactly two solutions $pmod{p_i^{e_i}}$.

Since each of the $k$ equation has two solutions, after CRT there will be $2^k$ base solutions. This in turn gives $2^k$ infinite classes.

Edit 1: If one of the primes is even then the combination of the odd primes gives $2^{k-1}$ classes. The total number depends on how many classes there are for $x^2equiv 0pmod{p_1^{e_1} = 2^{e_1}}$.

Since x and y have a difference like $d$,we can have a single variable quadratic equation by transformation.
If $x=y ±d$ we have:

$a(y ±d)^2+by+c=0$

$ay^2 +(b ±2ad)y ±ad^2+c=0$

$Delta=(b ±2ad)^2-4a(c ±ad^2)>=0$

Limitation of value of $Delta$ gives us a short range of numbers to try for solution. For example we solve $x^2+1=25y$:

We consider two cases:

$|x|>|y|$; so we can transform equation as follows:

Let $x= y+d$ , we have:

$(y+d)^2-25y+1=0$

$y^2+(2d-25)y+d^2+1=0$

$Delta=-100 d +621$

For a solution in set of real numbers we must have:

$Delta=-100 a +621>=0$ ⇒ $100a<=621$ ⇒$a<6$ .

Now we have a small range to try, for example :

$a=5$ ⇒$Delta=121$ ⇒ $y=2$ and $y=13$

$y=2 ⇒ x=7$

$y=13 ⇒ x=18$.

Now suppose $|x|<|y|$, we have:

$(y-d)^2+1-25y=0$

$y^2+(-2a-25)y+a^2+1=0$

Which gives $y=2, x=7$ and $y=41, x=32$

For a general formula for x and y, the homogeneous equation $xtimes x-25y=0$ gives $x=25$ and $y=1$ and we have:

$x=7$ as a single solution so the equation for x is $x=25t +7$.

Where t ∈ Z. For y we put x in equation:

$(25t+7)^2+1=25y$ which gives $y=25t^2+14t+2$.

Hence there are infinitely many solutions for this equation.