M32-08

Official Solution:


If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?

This is indeed a complex question that requires a meticulous approach and careful reading of the solution.

First of all, \(-x \leq y \leq x\) guarantees two things:

Moreover, \(-x \leq x\) implies that \(x \geq 0\).

And finally, before moving to the statements, let's rephrase the question:

(1) \(xy\) is NOT a square of an integer.

If \(y = 0\) was the case, then \(xy\) would be 0, which is a square of an integer. This contradicts the statement.

On the other hand, if \(x = -y\) and \(y ≠ 0\) were true, for instance if \(x = -y = 1\), then \(xy\) would be \(-x^2\), a negative value, which cannot be a square of an integer. So, \(x = -y\) IS a possibility.

However, there may be other solutions where \(x\) is not equal to \(-y\) and yet the statement could still hold true. For instance, \(x = 2\) and \(y =1\).

Thus, we have two different answers to the question: Not sufficient.

(2) Point \((x, y)\) is NOT on x-axis.

If \(y = 0\) was the case, then the point \((x, y)\) would be ON the x-axis. Therefore, we know that \(y ≠ 0\).

If \(x = -y\) and \(y ≠ 0\) were true, for instance if \(x = -y = 1\), then the point \((x, y)\) would be \((x, -x)\), so (positive, negative), which is a point below the x-axis. Thus, \(x = -y\) IS a possibility.

Similarly, \(x\) can be not equal to \(-y\), and the statement could still hold true. For example, \(x = 2\) and \(y =1\).

Thus, we have two different answers to the question: Not sufficient.

(1)+(2) We could still have the cases considered above: for instance, if \(x = -y\) and \(y ≠ 0\), such as when \(x = -y = 1\), the answer to our original question would be YES. However, if we consider a different case, like \(x = 2\) and \(y = 1\), the answer would be NO. Not sufficient.


Answer: E