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If p and q are integers and p^{2} - q^{2} is even, which of the following must be true? I. p and q have the same parity. II. p + q is even. III. pq is even.

Explanation	Calculations	Help
$We first note that " p^{2} - q^{2} is even" implies that both p squared and q squared are either even or odd. In either case, p and q share the same parity.$	$p^{2} - q^{2} = even If p^{2}, q^{2} are even then p, q are even If p^{2}, q^{2} are odd then p, q are odd$	Theory & Tactics Method Card TRANS5-A Click to view full details
$We then observe that adding two even integers or two odd integers always gives an even integer, so p plus q is even.$	$p + q = 2 k + 2 m = 2 (k + m) p + q = (2 k + 1) + (2 m + 1) = 2 (k + m + 1)$	Theory & Tactics Method Card OE1-B Click to view full details Theory & Tactics Method Card OE1-A Click to view full details
$Finally, we see that multiplying two even integers gives an even number while multiplying two odd integers gives an odd number, so the product of p and q is not necessarily even.$	$pq = 2 k \times 2 m = 4 km = 2 (2 km) pq = (2 k + 1) (2 m + 1) = 4 km + 2 k + 2 m + 1 = 2 (2 km + k + m) + 1$	Theory & Tactics Method Card OE2-B Click to view full details Theory & Tactics Method Card OE2-A Click to view full details

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D

If p and q are integers and p^{2} - q^{2} is even, which of the following must be true? I. p and q have the same parity. II. p + q is even. III. pq is even.

Explanation	Calculations	Help
$We first note that " p^{2} - q^{2} is even" implies that both p squared and q squared are either even or odd. In either case, p and q share the same parity.$	$p^{2} - q^{2} = even If p^{2}, q^{2} are even then p, q are even If p^{2}, q^{2} are odd then p, q are odd$	Theory & Tactics Method Card TRANS5-A Click to view full details
$We then observe that adding two even integers or two odd integers always gives an even integer, so p plus q is even.$	$p + q = 2 k + 2 m = 2 (k + m) p + q = (2 k + 1) + (2 m + 1) = 2 (k + m + 1)$	Theory & Tactics Method Card OE1-B Click to view full details Theory & Tactics Method Card OE1-A Click to view full details
$Finally, we see that multiplying two even integers gives an even number while multiplying two odd integers gives an odd number, so the product of p and q is not necessarily even.$	$pq = 2 k \times 2 m = 4 km = 2 (2 km) pq = (2 k + 1) (2 m + 1) = 4 km + 2 k + 2 m + 1 = 2 (2 km + k + m) + 1$	Theory & Tactics Method Card OE2-B Click to view full details Theory & Tactics Method Card OE2-A Click to view full details

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D