If x and y are integers, is (x-y)^2 > (x+y)^2?

Target question: Is \((x-y)^2 > (x+y)^2\)?
This is a good candidate for rephrasing the target question.

Given: Is \((x-y)^2 > (x+y)^2\)?
Expand and simplify both sides to get: Is \(x^2-2xy+y^2 > x^2+2xy +y^2\)?
Subtract \(x^2\) and \(y^2\) from both sides to get: Is \(-2xy > 2xy\)?
Add \(2xy\) to both sides to get: Is \(0 > 4xy\)?

REPHRASED target question: Is \(0 > 4xy\)?
Aside: the video below has tips on rephrasing the target question

Statement 1: \(|x|y > x|y|\)
First recognize that:
i) If x and y are both POSITIVE, then \(|x|y = x|y|\)
ii) If x and y are both NEGATIVE, then \(|x|y = x|y|\)
iii) If x is POSITIVE and y is NEGATIVE, then \(|x|y < x|y|\)
iv) If x is NEGATIVE and y is POSITIVE, then \(|x|y > x|y|\)

You can verify this by testing various values of x and y

Notice that case iv is the only case that satisfies the given inequality \(|x|y > x|y|\)
So, it must be the case that x is NEGATIVE and y is POSITIVE

This means that 4xy = (4)(NEGATIVE)(POSITIVE) = some negative value, which means 4xy is definitely less than zero
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: \(-1 < x^{-y} < 0\)
This tells us that \(x^{-y}\) is NEGATIVE, which means the base (x) must be NEGATIVE
Also since ODD exponents preserve of the sign of the base, we can also conclude that y is ODD

Also notice that, if y is NEGATIVE, then -y is POSITIVE
If x is a NEGATIVE integer, and -y is a POSITIVE odd integer, then \(x^{-y} < -1\)[/b]
HOWEVER, we're told that \(-1 < x^{-y} < 0\)[/b]
This means it must be the case that -y is NEGATIVE, which means y is positive

Now that we know that xis NEGATIVE and y is positive, we're in the exact same situation we were in that statement 1
That is, 4xy is definitely less than zero
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: D

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Cheers,
Brent