由 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……