Question

Prove that 6 divides (a + b + c) if and only if 6 divides (a
^3 + b^3 + c
^3
).​

Answer 1

Answer:

Step-by-step explanation:

Since  n3−n=(n−1)n(n+1)  is a product of  3  consecutive integers,  3!∣(n3−n) . Therefore, if  a1,…,am∈Z ,  6∣(a3i−ai) ,  i∈{1,…,m} . Hence  6∣((a31+⋯+a3m)−(a1+⋯+am) . So if  6∣(a1+⋯+am) , then  6∣(a31+⋯+a3m) ; and if  6∣(a31+⋯+a3m) , then  6∣(a1+⋯+am) .

We have shown that  6∣(a1+⋯+am)⇔6∣(a31+⋯+a3m) .