Find all integer solutions of $3(m^2 + n^2) - 7(m+n) = -4$.

Question

The solution and further information about how to solve this type of equation about how to solve this type of lattice point and circles will be much appreciated. Thanks in advance.

Answer 1

First, expand the equation and bring the $-4$ to the other side.

$$3m^2+3n^2-7m-7n+4 = 0$$

To solve for $m$, you can rearrange the LHS:

$$3m^2-7m+\left(3n^2-7n+4\right) = 0$$

By the Quadratic Formula,

$$m = \frac{-(-7)\pm\sqrt{(-7)^2-4(3)\left(3n^2-7n+4\right)}}{2(3)}$$

$$m = \frac{7\pm\sqrt{1-36n^2+84n}}{6}$$

You want the discriminant to be a perfect square:

$$1-36n^2+84n = t^2$$

and

$$1-36n^2+84n \geq 0$$

Answer 2

Hint:

$$3m^2-7m+3n^2-7n+4=0$$

What is the discriminant of the quadratic equation in $m$

Answer 3

Following the comment by @saulspatz, we get $$ (6m-7)^2+(6n-7)^2=50 $$ So, we have to solve $a^2+b^2=50$ with $a \equiv b \equiv -7 \bmod 6$.

The solutions are $$ \begin{array}{cccc} a & b & m & n \\ -7 & -1 & 0 & 1 \\ -1 & -7 & 1 & 0 \\ 5 & 5 & 2 & 2 \\ \end{array} $$ These were found by brute force. For numbers larger than $50$, consider its prime factorization and use the Brahmagupta–Fibonacci identity.