Problem 1697

1. We start with the given equation for strictly positive real numbers $a_1, a_2, \ldots, a_n$:
$$a_1^2 + a_2^2 + \cdots + a_n^2 = a_1 + a_2 + \cdots + a_n + \frac{n(n^2 + 6n + 11)}{3}$$
2. We rewrite the equation using summation notation:
$$\sum_{k=1}^n (a_k^2 - a_k) = \frac{n(n^2 + 6n + 11)}{3}$$
3. Consider the equation for $n-1$:
$$\sum_{k=1}^{n-1} (a_k^2 - a_k) = \frac{(n-1)((n-1)^2 + 6(n-1) + 11)}{3}$$
4. Subtract the equation for $n-1$ from the equation for $n$:
$$\sum_{k=1}^n (a_k^2 - a_k) - \sum_{k=1}^{n-1} (a_k^2 - a_k) = \frac{n(n^2 + 6n + 11)}{3} - \frac{(n-1)((n-1)^2 + 6(n-1) + 11)}{3}$$
5. Simplify the left-hand side:
$$a_n^2 - a_n = \frac{n(n^2 + 6n + 11)}{3} - \frac{(n-1)((n-1)^2 + 6(n-1) + 11)}{3}$$
6. Simplify the right-hand side:
$$\frac{n(n^2 + 6n + 11) - (n-1)((n-1)^2 + 6(n-1) + 11)}{3}$$
7. Expand and simplify the numerator:
$$n^3 + 6n^2 + 11n - ((n-1)^3 + 6(n-1)^2 + 11(n-1))$$
$$= n^3 + 6n^2 + 11n - (n^3 - 3n^2 + 3n - 1 + 6n^2 - 12n + 6 + 11n - 11)$$
$$= n^3 + 6n^2 + 11n - n^3 + 3n^2 - 3n + 1 - 6n^2 + 12n - 6 - 11n + 11$$
$$= 3n^2 + 9n + 6$$
8. Thus, we have:
$$a_n^2 - a_n = \frac{3n^2 + 9n + 6}{3} = n^2 + 3n + 2$$
9. Factorize the quadratic equation:
$$a_n^2 - a_n - (n^2 + 3n + 2) = 0$$
$$(a_n - (n + 2))(a_n + (n + 1)) = 0$$
10. Since $a_n$ is strictly positive, we discard the negative solution:
$$a_n = n + 2$$
11. Verify the solution by substituting $a_n = n + 2$ back into the original equation:
$$\sum_{k=1}^n (k+2)^2 = \sum_{k=1}^n (k+2) + \frac{n(n^2 + 6n + 11)}{3}$$
$$\sum_{k=1}^n (k^2 + 4k + 4) = \sum_{k=1}^n k + 2n + \frac{n(n^2 + 6n + 11)}{3}$$
$$\sum_{k=1}^n k^2 + 4\sum_{k=1}^n k + 4n = \sum_{k=1}^n k + 2n + \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1)}{6} + 4 \cdot \frac{n(n+1)}{2} + 4n = \frac{n(n+1)}{2} + 2n + \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1)}{6} + 2n(n+1) + 4n = \frac{n(n+1)}{2} + 2n + \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{3n(n+1) + 12n + 2n(n^2 + 6n + 11)}{6}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
$$\frac{n(n+1)(2n+1) + 12n(n+1) + 24n}{6} = \frac{n(n^2 + 6n + 11)}{3}$$
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