Manipulate me

n^2 + 3 n + 2 = 7 k

(n + 1)(n + 2) = 7 k

Then n + 1 is divisible by 7 , n + 2 is divisible by 7 so

n = 6 , 13 , 20 , .......... , 1987 (arithmetic progression of base 7 with 284 terms)

n = 5, 12 , 19 , .......... , 1986 (arithmetic progression of base 7 with 284 terms)

We have to prove that , no terms in common

Let the p th term in the first sequence equals the q th term in the second one

Then

6 + 7(p - 1) = 5 + 7(q - 1)

So

q - p = 1/7 (refused because p , q are integers)

So

The answer is 284 + 284 = 568

By comparing the possible remainders it is easy to see that $n\equiv 5(mod\quad 7)\quad or\quad n\equiv 6(mod\quad 7)$

So all numbers of the form $7a+5$ and $7a+6$ satisfy the condition.

Smallest number of the form $7a+5$ is $7\cdot 0+5$ and the biggest is $7\cdot 283+5$ so we have $284$ numbers.

Similarly the smallest number of te form $7a+6$ is $7\cdot 0+6$ and the biggest is $7\cdot 283+6$ so we have $284$ numbers.

Adding those values up we get the answer $568$ .