Official Solution:If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)? This is a hard question. You should pay attention to every detail and read the solution very carefully First of all, \(-x \leq y \leq x\) ensures two things:

1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.

2. \(x+y\geq 0\), so the square root won't be equal to negative number.

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

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

Does \(\sqrt{x^2 - y^2} = x + y\)?

Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?

Does \(xy+y^2=0\)?

Notice here that we cannot reduce this by \(y\), because we'll loose a possible root: \(y=0\). Does \(y(x+y)=0\)?

Does \(y=0\) or \(x=-y\)?

(1) \(xy\) is NOT a square of an integer. If \(y = 0\) were true, then \(xy\) would be 0, which is a square of an integer.

However, if \(x = -y\) and \(x ≠ 0\) were true, for example if \(x = -y = 1\), then \(xy\) would be \(-x^2\), so negative, which cannot be a square of an integer. So, \(x = -y\) IS possible.

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

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

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

If \(y = 0\) were true, then point \((x, y)\) would be ON the x-axis. So, we know that \(y ≠ 0\).

However, if \(x = -y\) and \(x ≠ 0\) were true, for example if \(x = -y = 1\), then then point \((x, y)\) would be \((x, -x)\), so (positive, negative), which would mean that it's below x-axis. So, \(x = -y\) IS possible.

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

(1)+(2) We could have the same cases: if \(x = -y\) and \(x ≠ 0\), for example if \(x = -y = 1\), then the answer is YES but if say \(x = 2\) and \(y =1\), then the answer is NO. Not sufficient.

Answer: E

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