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由 **kevin** 於星期日五月 11, 2003 11:35 am

轉貼自昌爸234fermat:
a=k(m^2-n^2)
b=2kmn
c=k(m^2+n^2)
2k^3 mn (m^2-n^2)(m^2+n^2)

Smallest case

2mn(m-n)(m^2+n^2)

if m or n =0(mod 2), then the product is divisible by 4.
Else both of them are odd, m-n =0(mod 2), the product is divisible by4.
If m or n=0(mod 3), the product is divisible by 3.
If m=n(mod 3), the product is divisible by 3.
If m=1, n=2(mod 3) or m=2 n=1(mod 3), 3｜m^2+n^2, the product is divisible
by 3.

Similar for proving divisible by 5

由 **kevin** 於星期六五月 10, 2003 7:50 pm

昌爸那邊的題目
請證明勾股數必為60的倍數
我看過siuhochung的解法.
我有更好的解法.只需因式分解即可
一起討論吧!

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[quote="kevin"]轉貼自昌爸234fermat: a=k(m^2-n^2) b=2kmn c=k(m^2+n^2) 2k^3 mn (m^2-n^2)(m^2+n^2) Smallest case 2mn(m-n)(m^2+n^2) if m or n =0(mod 2), then the product is divisible by 4. Else both of them are odd, m-n =0(mod 2), the product is divisible by4. If m or n=0(mod 3), the product is divisible by 3. If m=n(mod 3), the product is divisible by 3. If m=1, n=2(mod 3) or m=2 n=1(mod 3), 3｜m^2+n^2, the product is divisible by 3. Similar for proving divisible by 5[/quote]