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How do we determine integral solutions to the following equation:

$$324x^2-8676x + 56700 = y^2$$

Where $x$ and $y$ are positive integers.

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up vote 2 down vote accepted

This can be transformed into:

$$(18x-241)^2 - 1381 = y^2$$

Thus:

$$(18x-241)^2 -y^2 = 1381 \Rightarrow (18x + y - 241)(18x - y -241) = 1381$$

But $1381$ is a prime number, thus only has two divisors: $1$ and $1381$.

So we have two systems of linear equations:

$$\begin{cases} 18x + y -241 = \pm 1381 \\ 18x - y -241 = \pm 1 \end{cases} $$

or

$$\begin{cases} 18x + y -241 = \pm 1 \\ 18x - y -241 = \pm 1381 \end{cases} $$

(where if you chose the negative option in one line, you must chose the negative option in the other one and vice-versa)

I'll leave this linear equations to you. But if you want to check the solutions, there are only two:

$$x_1 = -25 \hspace{1cm} y_1 = 690 \hspace{1cm}\text{ and }\hspace{1cm} x_2 = -25 \hspace{1cm} y_2 =-690$$

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$\begin{eqnarray} {\bf Hint}\qquad\quad c &=\!& (ax)^2 - 2b(ax) -y^2\\ \iff\ \ b^2\!+c &=\!& (ax-b)^2-y^2 &&\text{by completing the square}\\ &=\!& (ax-b-y)(ax-b+y)&&\text{by difference of squares}\end{eqnarray}$

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