Let $ P(x) = ax^2+bx+c$. We are looking for the maximum number of integers $ n \ge 2$ such that: $ P(n+1) = P(n) + P(n-1)$ $ a(n+1)^2 + b(n+1) + c = a(n)^2+b(n)+c + a(n-1)^2 + b(n-1) + c$ $ (a)n^2 + (2a+b)n + (a+b+c) = (2a)n^2 + (2b-2a)n + (a-b+2c)$ $ (a)n^2 + (b-4a)n + (c-2b) = 0$ This is a quadratic which can have at most $ 2$ integer solutions $ n \ge 2$. $ (a,b,c) = (1,-1,4)$ yields $ n^2-5n+6 = 0 \Rightarrow n=2,3$. With $ P(x) = x^2-x+4$, we have: $ P(1) = 4$, $ P(2) = 6$, $ P(3) = 10$, and $ P(4) = 16$. So, $ P(3) = P(2)+P(1)$ and $ P(4) = P(3)+P(2)$. Thus, the maximum number of terms that are equal to the sum of the two preceding terms is $ 2$.