有關a^2+b^2=c^2

反證法 :設a , b 同時是奇數 :a = 2p + 1
b = 2k + 1則
a^2 + b^2 = c^2(2p+1)^2 + (2k+1)^2 = c^2(4p^2 + 4p + 1) + (4k^2 + 4k + 1) = c^24(p^2 + p + k^2 + k) + 2 = c^2故 c^2 是 偶數推得 c 也必為偶數,設 c = 2t ,則 c^2 = (2t)^2 = 4t^2 是 4 的倍數 ,這與 c^2 = 4(p^2 + p + k^2 + k) + 2 不是 4 的倍數茅盾。故a^2+b^2=c^2 中 a,b不可同時是奇數。

權威解答:
Since if a and b are both odd, then the left side must be mutiple of 2, implying
that c must be even too, the equation can be further simplified by divided by 2
to obtain the same form, P^2+Q^2=R^2, if we do this continuously ( to assume
P and Q are both odd), it will be meaningless, and doing the same things, so
we initially assume that a and b are not both odd.