Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do we determine integral solutions to the following equation:

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

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

share|cite|improve this question
up vote 2 down vote accepted

This can be transformed into:

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


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

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

So we have two systems of linear equations:

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


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

(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:

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

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.