难度：++
3，4，5 是最小的 Pythagoras 数组，而且 3*4*5 = 60。证明对任意 Pythagoras 数组 a，b，c，a*b*c 都能被60整除。
Let a2+b2=c2 , we need to prove 3, 4, 5 divide abc. i) To prove 4 divides abc: 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: 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. |
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