YLL討論網

WLOG, we can let gcd(a,b,c,d)=1#ed_op#br#ed_cl#a+b+c+d = lcm(a,b,c,d)#ed_op#br#ed_cl##ed_op#br#ed_cl#WLOG, let a,b,c <= d#ed_op#br#ed_cl#a+b+c+d  < 4d#ed_op#br#ed_cl#a+b+c  < 3d#ed_op#br#ed_cl##ed_op#br#ed_cl#a+b+c+d = kd  (k>1)#ed_op#br#ed_cl##ed_op#br#ed_cl#So k = 2.#ed_op#br#ed_cl##ed_op#br#ed_cl#lcm(a,b,c,d) = 2d#ed_op#br#ed_cl##ed_op#br#ed_cl#a,b,c should be factors of d or twice the factors of d.#ed_op#br#ed_cl##ed_op#br#ed_cl#a+b+c+d = 2d#ed_op#br#ed_cl#a+b+c = d#ed_op#br#ed_cl##ed_op#br#ed_cl#As gcd(a,b,c,d)=1 , gcd(a,b,c)=1.#ed_op#br#ed_cl#From a+b+c = d & gcd(a,b,c)=1, a,b and c are pairwise coprime.#ed_op#br#ed_cl##ed_op#br#ed_cl#Let d is odd.#ed_op#br#ed_cl#Considering parity, excatly two of a,b,c are twice the factors of d.#ed_op#br#ed_cl#Let b,c be those number and the factors of d be x,y.#ed_op#br#ed_cl#WLOG, let x >= y#ed_op#br#ed_cl##ed_op#br#ed_cl#a + 2 x + 2 y = d#ed_op#br#ed_cl#a + 4 x >= d#ed_op#br#ed_cl#5x >= d or 5a >= d#ed_op#br#ed_cl#But a or x != d, d mod 2 = 1#ed_op#br#ed_cl#3a or 3x or 5x or 5a = d#ed_op#br#ed_cl##ed_op#br#ed_cl#It shows that 3 or 5|abcd.#ed_op#br#ed_cl##ed_op#br#ed_cl#Let d is even. 沒空證……

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。