Solution of the Diophantine Equation $x^2-ay+c=0.$

Question

I am just wondering if there is a general method in solving the Diophantine equation

$x^2-ay+c=0$,

where $a$ and $c$ are positive integers. This question arose because I need to find a solution to the equation $125+x^2=22y$. I am so thinking that trial and error is not a suitable method to do.

Thanks for your help.

Answer 1

I will solve just one specific case:

For this case, we need to solve $x^2 \equiv -125 \pmod{22}$

We have $22|(x^2+125)\Rightarrow22|(x^2-7+132)\Rightarrow22|(x^2-7)$ because $132$ is divisible by $22$.

So now we need to solve $x^2 \equiv 7 \pmod{22}$

Trial and error is actually suitable, but only if you notice that because $x^2-7$ is divisible by $22$, $x$ must be an odd number. This will reduce the number of cases from $22$ to $11$. With $11$ cases, we can easily set up a table like this:

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Remainder of }x\div22 &1&3&5&7&9&11&13&15&17&19&21\ \\ \hline \text{Remainder of }x^2\div22 &1&9&3&5&15&11&15&5&3&9&1\\\hline \text{Remainder of }(x^2-7)\div22 &16&2&18&20&8&4&8&20&18&2&16\\\hline \end{array}

Based on the table above, there are no positive integers $x$ satisfy $x^2 \equiv 7 \pmod{22}$, so there are no positive integers $x,y$ overall satisfy $125+x^2=22y$.

Answer 2

Above equation shown below:

$x^2-ay+c=0$ -----------$(1)$

Equation $(1)$ has parametric solution for $(a,c)= (2,9)$

Where $(x,y)= ((2k+9), (2k^2+18k^2+45))$

For k=4, we get:

$(17) ^2-2(149)+9=0$