Question 475377
By the Pythagorean theorem,


*[tex \LARGE a^2 + b^2 = c^2]


Assume c is even. Then c^2 ≡ 0 (mod 4), and a^2 or b^2 can only be congruent to 0 or 1 mod 4, so this creates a contradiction (unless a,b,c are all divisible by 2, which still doesn't work since the triple is primitive). Hence, c must be odd.