quadratic diophantine's equation in form of $y=ax^2+bx+c$ I stumbled on this on Geogebra. Actually i would like to set integers pair $x$ $y$ that fits the general quadratic form.  
Given $(x_1,y_1)$ and $(x_2,y_2)$ are integers pairs, i am looking for set of possible solution of integers $a,  b$ and $c$ in $$y=ax^2+bx+c$$
Please help. 
 A: Assuming $x_2 > x_1$, to simplify reasoning, we have:
$$
\begin{matrix}
x_1^2 A + x_1 B + C = y_1 \\
x_2^2 A + x_2 B + C = y_2 
\end{matrix}
\quad (*) 
$$
Taking the difference one notes that the integer unknowns $A$, $B$ are subject to a linear Diophantine equation of the type
$$
  a A + b B = c
$$
here
$$
(x_2^2 - x_1^2) A + (x_2 - x_1) B = (y_2 - y_1) \quad (**) \\
$$
For such an equation $(**)$, a solution algorithm exists.
To have solutions 
\begin{align}
d &= \gcd(a,b) \\
  &= \gcd(x_2^2 - x_1^2, x_2 - x_1) \\
  &= \gcd((x_2 + x_1)(x_2 - x_1), x_2 - x_1) \\
  &= \lvert x_2 - x_1 \rvert \gcd(x_1 + x_2, 1) \\
  &= \lvert x_2 - x_1 \rvert \\
  &= x_2 - x_1
\end{align}
must be a divisor of $c = y_2 - y_1$.
$$
k := \frac{c}{d} = \frac{y_2 - y_1}{x_2 - x_1} \in \mathbb{Z}
$$ 
The equation
$$
a s + b t = (x_2 - x_1)(x_1 + x_2) s + (x_2 - x_1) t = d = x_2 - x_1
$$
can be solved by
$$
s = 0 \quad t = 1
$$
so we have the particular solution
$$
A_0 = 0 \quad B_0 = \frac{y_2-y_1}{x_2 - x_1} = k
$$
and the general solution
$$
A = t \quad B = k - t (x_1 + x_2) \quad (t \in \mathbb{Z})
$$
for equation $(**)$.
Now we put that solution into system $(*)$ and solve for $C$:
$$
\begin{align}
C &= y_1 - x_1^2 t + x_1\left( t (x_1 + x_2) - k \right)
= y_1 - x_1 k + x_1 x_2 t
 \\ 
C &= y_2 - x_2^2 t + x_2\left( t (x_1 + x_2) - k \right)
= y_2 - x_2 k + x_1 x_2 t
\end{align} 
$$
We need both $C$ values to be the same, which gives
$$
y_1 - x_1 k + x_1 x_2 t = y_2 - x_2 k + x_1 x_2 t \iff \\
(x_2 - x_1) k = y_2 - y_1
$$
which is independent of $t$ and fits the definition of $k$.
Summary:
If 
$$
k = \frac{y_2 - y_1}{x_2 - x_1} \in \mathbb{Z}
$$
we have infinite many solutions
$$
A = t \quad B = k - t(x_1 + x_2) \quad
C =  y_1 - x_1 k + x_1 x_2 t
\quad (t \in \mathbb{Z})
$$
for the system of equations $(*)$.
Test:
$$
x_1^2 t + x_1(k-t(x_1 + x_2)) + y_1 - x_1 k + x_1 x_2 t = y_1 \\
x_2^2 t + x_2(k-t(x_1 + x_2)) + y_1 - x_1 k + x_1 x_2 t = 
(x_2 - x_1) k + y_1 = y_2 - y_1 + y_1 = y_2
$$
A: Subtracting the two equations memberwise,
$$(a(x_1+x_2)+b)(x_1-x_2)=y_1-y_2.$$
Then if $x_1-x_2$ divides $y_1-y_2$,
$$a(x_1+x_2)+b=q,$$
$$b=q-a(x_1+x_2),$$
and from the first equation,
$$c=y_1-ax_1^2-bx_1.$$
Now if you want to find a third pair that fits the model,
$$0=ax_3^2+bx_3+c-y_3=ax_3^2+(q-a(x_1+x_2))x_3+y_1-ax_1^2-(q-a(x_1+x_2))x_1-y_3\\
=(x_1-x_3)(x_2-x_3)a-q(x_1-x_3)+y_1-y_3,\\
=(x_1-x_3)((x_2-x_3)a-q)+y_1-y_3.$$
It requires $x_1-x_3$ to divide $y_1-y_3$, then $x_2-x_3$ to divide $q-q'$.
A: One can solve the system of linear Diophantine equations in $a,b,c$
$$
y_1 = ax_1^2 + bx_1 + c \\ y_2 = ax_2^2 + bx_2 + c;
$$
there are algorithms to do this, e.g., see here. However, over $\mathbb{Z}$ it is certainly more difficult than in the case over a field $K$. Of course, for special cases it can be very easy, e.g., for $(x_1,y_1)=(0,0)$ and $(x_2,y_2)=(1,1)$ it gives $c=0$ and all $a,b\in \mathbb{Z}$ with $a+b=1$.
