First we will rearrange the equation so that it looks like a quadratic equation in x.

x² – (4y² + 8y)x + 4 = 0

Now we can find an expression for x in terms of y, using the quadratic equation formula.

x = [(4y² + 8y) +/- √{(4y² + 8y)² – 4² }]/2

Now for some simplifying!

x = (4y² + 8y)/2 +/- [√{(4y² + 8y – 4)( 4y² + 8y + 4}]/2

x = (2y² + 4y) +/- [2√{(4y² + 8y – 4)( y² + 2y + 1)}]/2

x = (2y² + 4y) +/- √{(4y² + 8y – 4)( y + 1)²}

x = (2y² + 4y) +/- 2(y + 1)√(4y² + 8y – 4)

x = (2y² + 4y) +/- 4(y + 1)√(y² + 2y – 1)

Now, this can only be an integer when y is an integer if y² + 2y – 1 is a square number, or, if y² + 2y – 1 is not a square number, but is multiplied by zero in the expression. Suppose it is a square number, then

y² + 2y – 1 = p²

y² + 2y + 1 = p² + 2

(y + 1)² – p² = 2

But… the difference between two square numbers cannot be two! The closest square numbers are 1 and 4, which have a difference of 3. Therefore y² + 2y – 1 cannot be a square number.

Therefore our only option is that y + 1 (which multiplies √(y² + 2y – 1) in our expression for x) is zero.

So we have y = -1, which means that x = 2 x 1 + 4 x -1 = -2.

There is only one integer solution, when (x, y) = (-2, -1).

Note: if you try graphing the equation using something like Desmos, you will see that (-2, -1) is not shown in the graph! Can you work out why?

Challenge 326: Integer Insanity