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

Explanation	Calculations	Help
$We first note that "m + n is odd" only when one integer is odd and the other is even. We translate "m + n is odd" accordingly.$	$m = 2 a + 1 n = 2 b m = 2 a n = 2 b + 1$	Theory & Tactics Method Card TRANS5-A Click to view full details
$For statement I we see that one integer is even and the other is odd, so their product is even.$	$mn = odd \times even = even mn = even \times odd = even$	Theory & Tactics Method Card OE2-C Click to view full details Theory & Tactics Method Card OE2-D Click to view full details
$For statement II, squaring preserves parity, and then adding an odd integer and an even integer yields an odd result.$	$m^{2} = odd \times odd = odd n^{2} = even \times even = even m^{2} + n^{2} = odd + even = odd$	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
$For statement III we subtract an odd integer and an even integer which gives an odd integer, and squaring that odd integer yields an odd result, so the expression is not even.$	$m - n = odd - even = odd m - n = even - odd = odd (m - n)^{2} = odd \times odd = odd$	Theory & Tactics Method Card OE3-C Click to view full details Theory & Tactics Method Card OE3-D Click to view full details Theory & Tactics Method Card OE2-A Click to view full details

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C

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

Explanation	Calculations	Help
$We first note that "m + n is odd" only when one integer is odd and the other is even. We translate "m + n is odd" accordingly.$	$m = 2 a + 1 n = 2 b m = 2 a n = 2 b + 1$	Theory & Tactics Method Card TRANS5-A Click to view full details
$For statement I we see that one integer is even and the other is odd, so their product is even.$	$mn = odd \times even = even mn = even \times odd = even$	Theory & Tactics Method Card OE2-C Click to view full details Theory & Tactics Method Card OE2-D Click to view full details
$For statement II, squaring preserves parity, and then adding an odd integer and an even integer yields an odd result.$	$m^{2} = odd \times odd = odd n^{2} = even \times even = even m^{2} + n^{2} = odd + even = odd$	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
$For statement III we subtract an odd integer and an even integer which gives an odd integer, and squaring that odd integer yields an odd result, so the expression is not even.$	$m - n = odd - even = odd m - n = even - odd = odd (m - n)^{2} = odd \times odd = odd$	Theory & Tactics Method Card OE3-C Click to view full details Theory & Tactics Method Card OE3-D Click to view full details Theory & Tactics Method Card OE2-A Click to view full details

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C