quadratic diophantine's equation in form of $y=ax^2+bx+c$

Question

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.

Answer 1

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 $q-q'$.

Answer 2

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 $a+b=1$.

Answer 3

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: $$ x_1^2 t + x_1(k-t(x_1 + x_2)) + y_1 - x_1 k + x_1 x_2 t = y_1 \\ x_2^2 t + x_2(k-t(x_1 + x_2)) + y_1 - x_1 k + x_1 x_2 t = (x_2 - x_1) k + y_1 = y_2 - y_1 + y_1 = y_2 $$