OFFICIAL SOLUTION:
If x and y are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?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. So, we know that y ≠ 0.
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-axisIf 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.
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