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If x and y are integers and x - y is odd, which of the following must be true? I. x y is even. II. x^{2} + y^{2} is odd. III. (x + y)^{2} is even.

Explanation	Calculations	Help
$We first note that x - y is odd only when one of x and y is odd and the other is even. We translate "x - y is odd" accordingly.$	$x is odd and y is even x is even and y is odd$	Theory & Tactics Method Card OE3-C Click to view full details Theory & Tactics Method Card OE3-D Click to view full details
$For statement I, since one of x and y is even and the other is odd, their product must be even.$	$x y = odd \times even = even x y = even \times odd = even$	Theory & Tactics Method Card OE2-C Click to view full details Theory & Tactics Method Card OE2-D Click to view full details
$For statement II, squaring preserves parity, and then adding an odd and an even yields an odd result.$	$x^{2} = odd \times odd = odd y^{2} = even \times even = even x^{2} + y^{2} = odd + even = odd$	Theory & Tactics Method Card OE2-A Click to view full details Theory & Tactics Method Card OE2-B Click to view full details Theory & Tactics Method Card OE1-C Click to view full details
$For statement III, adding an odd and an even gives an odd, and squaring an odd gives an odd, so (x + y)^{2} is odd, not even.$	$x + y = odd + even = odd (x + y)^{2} = odd \times odd = odd$	Theory & Tactics Method Card OE1-C Click to view full details Theory & Tactics Method Card OE2-A Click to view full details

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I and II only

If x and y are integers and x - y is odd, which of the following must be true? I. x y is even. II. x^{2} + y^{2} is odd. III. (x + y)^{2} is even.

Explanation	Calculations	Help
$We first note that x - y is odd only when one of x and y is odd and the other is even. We translate "x - y is odd" accordingly.$	$x is odd and y is even x is even and y is odd$	Theory & Tactics Method Card OE3-C Click to view full details Theory & Tactics Method Card OE3-D Click to view full details
$For statement I, since one of x and y is even and the other is odd, their product must be even.$	$x y = odd \times even = even x y = even \times odd = even$	Theory & Tactics Method Card OE2-C Click to view full details Theory & Tactics Method Card OE2-D Click to view full details
$For statement II, squaring preserves parity, and then adding an odd and an even yields an odd result.$	$x^{2} = odd \times odd = odd y^{2} = even \times even = even x^{2} + y^{2} = odd + even = odd$	Theory & Tactics Method Card OE2-A Click to view full details Theory & Tactics Method Card OE2-B Click to view full details Theory & Tactics Method Card OE1-C Click to view full details
$For statement III, adding an odd and an even gives an odd, and squaring an odd gives an odd, so (x + y)^{2} is odd, not even.$	$x + y = odd + even = odd (x + y)^{2} = odd \times odd = odd$	Theory & Tactics Method Card OE1-C Click to view full details Theory & Tactics Method Card OE2-A Click to view full details

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I and II only