It is obvious that either all three are even, or only one is even.
If all a,b,c are even, done.
If c is even, a,b are odd, the left side (a2+b2) mod 4 is 2, but the right c2 mod 4 is 0, impossible.
If a (similarly for b) is even, b,c are odd, let b=2m+1, c=2n+1, then
a2=c2-b2=4(n-m)(n+m+1), (n-m) or (n+m+1) is even (as their difference is odd), so a is divisible by 4.
ii) To prove 3 divides abc:www.ddhw.com
Suppose there is no one divisible by 3, then all a2, b2, c2 mod 3 are 1,
1+1=1(mod 3), contradiction.
iii) To prove 5 divides abc
Suppose there is no one divisible by 5, then all a2, b2, c2 mod 5 are 1 or 4,
No matter how to arrange 1, 4, it is impossible to obtain an equation of a2+b2=c2 mod 5, contradiction.