Show that if a*n^2 + b*n + c (a, b, and c are integers)is divisible by seven for three consecutive integers, then it is divisible by seven for all integers.

Tautology; that which is, by definition, tautological.

A more tedious method: there are 3 numbers such that 7 divides (an^2+bn+c), and (a(n+1)^2+b(n+1)+c) , and (a(n+2)^2+b(n+2)+c). We expand the latter two and deduct the common part, (an^2+bn+c), giving: 2ax+a+b, 4ax+4a+2b. Subtracting twice the smaller from the larger gives 2a=7k, so 7 divides a. But if so, then bn+c=7l, bn+b+c=7m, and, again by subtraction, 7 divides b, too. But if so, 7 must also divide c.

Edited on March 11, 2012, 3:23 am

Posted by broll on 2012-03-11 01:34:26