Solving the Integer Equation (x - 1)² + (y - 2)² = (x – 1) (y - 2)

We are given the equation \(\left(x - 1\right)^2 + \left(y - 2\right)^2 = \left(x - 1\right)\left(y - 2\right)\), where \(x\) and \(y\) are integers. Our goal is to find the values of \(x\) and \(y\) that satisfy this equation and then calculate the value of \(2x + 3y\).

Simplifying the Given Equation

Let's simplify the equation by making a substitution to make it easier to work with.

• Let \(a = x - 1\).
• Let \(b = y - 2\).

Since \(x\) and \(y\) are integers, \(a\) and \(b\) must also be integers.

The equation now becomes:

\(\qquad a^2 + b^2 = ab\)

Rearranging the Equation

To solve this equation for integers \(a\) and \(b\), let's rearrange it:

\(\qquad a^2 - ab + b^2 = 0\)

This equation looks like a quadratic form. We can try to complete the square or manipulate it to find the integer solutions.

Let's multiply the entire equation by 2:

\(\qquad 2a^2 - 2ab + 2b^2 = 0\)

We can rewrite the left side by grouping terms to form squares:

\(\qquad (a^2 - 2ab + b^2) + a^2 + b^2 = 0\)

Recognize that the term in the parenthesis is a perfect square, \((a - b)^2\).

So, the equation becomes:

\(\qquad (a - b)^2 + a^2 + b^2 = 0\)

Finding Integer Solutions for a and b

We have the sum of three squared terms equal to zero: \((a - b)^2\), \(a^2\), and \(b^2\).

• For any real numbers \(a\) and \(b\), \((a - b)^2 \ge 0\), \(a^2 \ge 0\), and \(b^2 \ge 0\).
• The sum of non-negative numbers can only be zero if each individual term is zero.

Therefore, for the equation \((a - b)^2 + a^2 + b^2 = 0\) to hold, we must have:

• \((a - b)^2 = 0 \implies a - b = 0 \implies a = b\)
• \(a^2 = 0 \implies a = 0\)
• \(b^2 = 0 \implies b = 0\)

These conditions are consistent. If \(a=0\) and \(b=0\), then \(a=b\) is also true (\(0=0\)).

So, the only real solution is \(a = 0\) and \(b = 0\). Since \(a\) and \(b\) must be integers, \(a=0\) and \(b=0\) is also the only integer solution.

Finding the Values of x and y

Now substitute back the original expressions for \(a\) and \(b\):

• \(a = x - 1 = 0 \implies x = 1\)
• \(b = y - 2 = 0 \implies y = 2\)

We have found that \(x = 1\) and \(y = 2\). These are integers, so they satisfy the given condition.

Calculating the Value of 2x + 3y

Finally, we need to find the value of \(2x + 3y\) using \(x=1\) and \(y=2\).

\(\qquad 2x + 3y = 2(1) + 3(2)\)

\(\qquad 2x + 3y = 2 + 6\)

\(\qquad 2x + 3y = 8\)

Thus, the value of \(2x + 3y\) is 8.

Revision Table: Key Concepts in Solving the Equation

Additional Information: Understanding the Quadratic Form \(a^2 - ab + b^2 = 0\)

The equation \(a^2 - ab + b^2 = 0\) is a specific type of homogeneous quadratic equation. While we solved it by completing the square, another way to see why \(a=0, b=0\) is the only real solution is by considering it as a quadratic in \(a\) (assuming \(b \neq 0\)).

Divide the equation \(a^2 - ab + b^2 = 0\) by \(b^2\) (assuming \(b \neq 0\)):

\(\qquad \frac{a^2}{b^2} - \frac{ab}{b^2} + \frac{b^2}{b^2} = 0\)

\(\qquad \left(\frac{a}{b}\right)^2 - \left(\frac{a}{b}\right) + 1 = 0\)

Let \(z = \frac{a}{b}\). The equation becomes:

\(\qquad z^2 - z + 1 = 0\)

We can find the solutions for \(z\) using the quadratic formula \(z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)}\):

\(\qquad z = \frac{1 \pm \sqrt{1 - 4}}{2}\)

\(\qquad z = \frac{1 \pm \sqrt{-3}}{2}\)

The solutions for \(z\) are complex numbers: \(z = \frac{1 \pm i\sqrt{3}}{2}\). These are the complex cube roots of -1 (specifically, \(e^{i\pi/3}\) and \(e^{-i\pi/3}\)).

Since \(z = \frac{a}{b}\) must be a real number (as \(a\) and \(b\) are real), there is no real value of \(z\) that satisfies \(z^2 - z + 1 = 0\). This implies that our initial assumption \(b \neq 0\) must be false. Therefore, \(b\) must be 0.

If \(b = 0\), substituting into \(a^2 - ab + b^2 = 0\) gives \(a^2 - a(0) + 0^2 = 0\), which simplifies to \(a^2 = 0\). This means \(a = 0\).

This confirms that the only real solution (and thus the only integer solution) to \(a^2 - ab + b^2 = 0\) is indeed \(a=0\) and \(b=0\).