[數論]數論競賽題6

[數論]數論競賽題6

宇智波鼬 於 星期三 十月 05, 2005 9:36 pm


設a ,b ,c ,d 為四個自然數,它們的最小公倍數為 a+b+c+d。試證:此四數之乘積abcd 可被3或5整除。
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宇智波鼬

 
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passby 於 星期五 六月 09, 2006 8:47 pm


WLOG, we can let gcd(a,b,c,d)=1
a+b+c+d = lcm(a,b,c,d)

WLOG, let a,b,c <= d
a+b+c+d  < 4d
a+b+c  < 3d

a+b+c+d = kd  (k>1)

So k = 2.

lcm(a,b,c,d) = 2d

a,b,c should be factors of d or twice the factors of d.

a+b+c+d = 2d
a+b+c = d

As gcd(a,b,c,d)=1 , gcd(a,b,c)=1.
From a+b+c = d & gcd(a,b,c)=1, a,b and c are pairwise coprime.

Let d is odd.
Considering parity, excatly two of a,b,c are twice the factors of d.
Let b,c be those number and the factors of d be x,y.
WLOG, let x >= y

a + 2 x + 2 y = d
a + 4 x >= d
5x >= d or 5a >= d
But a or x != d, d mod 2 = 1
3a or 3x or 5x or 5a = d

It shows that 3 or 5|abcd.

Let d is even. 沒空證……

passby
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