The condition $a_{i+1} - 2a_i + a_{i-1} = a^2_i$ for $i = 1, 2, \ldots, N - 1$ is a herring; it is enough to know $a_{i+1} - 2a_i + a_{i-1} \geq 0$. Consider $x_i = \textrm{e}^{-a_i}$; then $x_1^2 = \textrm{e}^{-2a_i} \geq \textrm{e}^{-a_{i-1} - a_{i+1}} = x_{i-1}x_{i+1}$, i.e. the sequence $(x_i)_{i=0}^{i=N}$ is log-concave, and as such, it is known (and easy to prove) it is unimodal, i.e. for some $0\leq j \leq N$ we have $x_0\leq x_1\leq \cdots \leq x_{j-1} \leq x_j \geq x_{j+1} \geq \cdots \geq x_N$. But that means $0 = a_0 \geq a_1\geq \cdots \geq a_{j-1} \geq a_j \leq a_{j+1} \leq \cdots \leq a_N = 0$.

mavropnevma wrote: Consider $x_i = \textrm{e}^{-a_i}; \dotsc$ Why so complicated? Just assume the maximum of the sequence would be at some $i$ not equal to $0$ or $N$. Then $a_i \ge a_{i+1}$ and $a_i \ge a_{i-1}$ which gives $0 \ge a_{i-1}+a_{i+1}-2a_i=a_i^2$ and thus $a_i=0$ which gives in every case the result that the maximum of the sequence is $0$.

Tintarn wrote: Why so complicated? Just because I wanted to bring into picture the wide-spread literature on log-convex and log-concave sequences; of course, the whole thing is trivial.

$a^2_i+a^2_{i-1}+a^2_{i-2}+...+a^2_2+a^2_1 = -(a_i+a_1)$ LHS is nonnegative, then RHS must nonnegative too. We are done (cmiiw)

Suppose $a_i >0.$ Then, $2a_i+a_i^2=a_{i-1}+a_{i+1} >0.$ Then, either $a_{i-1}$ or $a_{i+1}$ is less than $0.$ We can continue using this process until we get $a_0$ or $a_N$ equal to $0,$ which is contradiction. Therefore, $a_i<0$ for all $1 \leq i \leq N-1.$

Claim 1. Suppose $a_{i+1}> 0$, $a_{i+1}-a_i>0$ for some $n-2\geq i\geq 0$. Then, $a_{i+2}> 0$ and $a_{i+2}-a_{i+1}>0$. Proof. We have $$a_{i+2}=a_{i+1}^2+2a_{i+1}-a_{i}> 0$$and $$a_{i+2}-a_{i+1}=a_{i+1}^2+a_{i+1}-a_{i}> 0.$$ Claim 2. Suppose $a_{i-1}\leq 0$ for some $n-1\geq i\geq 1$. Then, $a_i\leq 0$. Proof. Suppose otherwise, i.e. $a_i>0$ and $a_{i-1}\leq 0$, hence $$a_{i+1}=a_i^2+2a_i-a_{i-1}> 0$$and $a_{i+1}-a_i=a_i^2+a_i-a_{i-1}>0$. By the first claim we have that $a_n>0$, which is a contradiction. Now by the second claim, we are done.

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