Find number of all pairs of real numbers (x, y) for whichx2 + y2 + xy = 133

Sure, let's solve the equation step by step.

The given equation is x² + y² + xy = 133.

We can rewrite this equation in the form of a perfect square. To do this, we add and subtract (1/4) inside the equation.

So, the equation becomes x² + xy + (1/4) - (1/4) + y² = 133.

This can be written as (x + y/2)² - (1/4) = 133.

Solving further, we get (x + y/2)² = 133 + (1/4) = 532/4 = 133.

Taking square root on both sides, we get x + y/2 = ± √133.

So, the pairs of real numbers (x, y) are those for which x = √133 - y/2 and x = -√133 - y/2.

There are infinitely many such pairs of real numbers (x, y).