(4) If m is odd, then m^2 + 1 is even.
Proof. Assume that m is odd. Then m = 2k + 1 for some integer k. So
m^2 + 1 = (2k +1)^2 + 1
= 4k^2 + 4k + 1 + 1
= 4k^2 + 4k +2
= 2(2k^2 + 2k + 1)
= 2q
where q = 2k^2 + 2k + 1 is an integer. So m^2 + 1 is even.
QED
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