If $ \sum a_i^2\le n^3$ then, we have $ n\ge\sqrt{\frac{\sum a_i^2}{n}}\ge\frac{\sum|a_i|}{n}\ge n$. Thus, we must have $ a_1=a_2=\cdots =a_n$ which has solution $ a_i=n$ for all $ n$. Otherwise, $ \sum a_i^2= n^3+1$. Suppose $ \sum a_i\ge n^2+1$. Then, we have $ \sqrt{n^2+\frac{1}{n}}=\sqrt{\frac{\sum a_i^2}{n}}\ge\frac{\sum|a_i|}{n}\ge n+\frac{1}{n}$ which is clearly false. Thus, $ \sum a_i=n^2$. It is well known that $ k\equiv k^2\pmod{2}$ for all integer $ k$. Thus, we have $ n^3+1=\sum a_i^2\equiv\sum a_i=n^2\pmod{2}$. That is, we have $ n^3+1\equiv n^2\pmod{2}$, an impossibility. Thus, the only solution is $ a_i=n$ for all $ i$.

$d_n=a_n-n$.Then $\sum^{n}_{i=1}a_i=\sum^{n}_{i=1}d_i+n^2$ and $\sum^{n}_{i=1}d_i\geq\0$ so $\sum^{n}_{i=1}a_i^2 =\sum^{n}_{i=1}d_1^2+2n\sum^{n}_{i=1}d_i+n^3\leq n^3+1$. We have $\sum^{n}_{i=1}d_i^2\leq 1$. $d_i$ is an integer we have $d_1=d_2=...=d_n=0$ or $d_1=d_2=d_{i-1}=d_{i+1}=...=d_n=0 |d_i|=0 $ . Answer: $a_i=n$ or $a_1=a_2=2=a_{i-1}=a_{i+1}=...=a_n$ and $a_i= n-1$ or $n+1$ There $n$ and $1\leq i\n$ $i$ are positive integer.

Is my solution correct? Can someone help me? This is what I have so far: Using RMS-AM inequality, we have \[ \frac{n^2-a_i}{n-1} \leq \frac{(a_1+a_2+\cdots a_n)-a_i}{n-1} \leq \sqrt{\frac{(a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^2)-a_i^2}{n-1}} \leq \sqrt{\frac{(n^3+1-a_{i}^2)}{n-1}} \] Expanding and simplifying, we obtain \[ n\cdot a_n^2-2n^2\cdot a_n+n^3-n+1\leq0 \] This implies that \[ n-\sqrt{\frac{n-1}{n}} \leq a_{i} \leq n+\sqrt{\frac{n-1}{n}} \] Since $a_i$ is integer, we conclude that $a_i=n$ for all $i$.

anything-- wrote: \[ \frac{(a_1+a_2+\cdots a_n)-a_i}{n-1} \leq \sqrt{\frac{(a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^2)-a_i^2}{n-1}} \] I think it is wrong!

LoveMath4ever wrote: anything-- wrote: \[ \frac{(a_1+a_2+\cdots a_n)-a_i}{n-1} \leq \sqrt{\frac{(a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^2)-a_i^2}{n-1}} \] I think it is wrong! no, it is true because it is equivalnet to: $(n-1)(\sum_{j\neq i}{a_j}^2) \ge (\sum_{j\neq i}{a_j})^2$ which is true by cauchy.

Without loss of generality, let $a_n \ge a_{n-1} \ge \cdots \ge a_1$. Let $\sum_{i = 1}^{n-1}a_i = A_1$ and $\sum_{i = 1}^{n-1}a_i^2 = A_2$. By Cauchy-Schwarz, $$A_1^2 = \left(\sum_{i = 1}^{n-1}a_i \cdot 1\right)^2 \le (n-1)A_2.$$By the given inequalities, we find that $$(n-1)(n^3 + 1 - a_n^2) \ge (n-1)A_2 \ge A_1^2 \ge (n^2 - a_n)^2.$$Expanding out both sides, we see that $$n^4 + n - na_n^2 - n^3 - 1 + a_n^2 \ge n^4 - 2n^2a_n + a_n^2.$$Rearranging both sides, we find that $$2n^2a_n \ge n^3 + 1 + na_n^2 - n.$$By the definition of $a_n$ and the given inequalities, we may let $a_n = b + n$ where $b \ge 0$ is an integer. Then, we have that $$2n^2(b+n) \ge n^3 + 1 + n(b+n)^2 - n = 2n^3 + 2bn^2 + bn - n + 1.$$Simplifying further, we see that $$0 \ge bn - n + 1.$$Thus, we have that $b = 0$. Since $a_n$ is the maximum $a_i$ and the sum of all the $a_i$ is greater than or equal to $n^2$, we must then have that all the $a_i$ are $n$ since we would otherwise have that $\sum_{i = 1}^n a_i < n^2$. Thus, the number of ordered $n$-tuples is $1$ for all $n$.

Just consider the sum of squares $S:=\sum_{i=1}^n (a_i-n)^2$. Then by using (1) and (2) we get $0 ~\le~ S ~\le~ (\sum_{i=1}^na_i^2) -2n(\sum_{i=1}^na_i) +\sum_{i=1}^nn^2 ~\le (n^3+1)-2n(n^2)+n\cdot n^2 ~=~ 1$. This means that at least $n-1$ of the squares $(a_i-n)^2$ are $0$, while the last one is either $0$ or $1$. $\bullet\,$ If all the squares are $0$, this yields the feasible solution $(n,n,n,\ldots,n)$. $\bullet\,$ If $n-1$ of the squares are $0$ and the last one is $1$, we get that $n-1$ of the $a_i$ $~~~~~$ are equal to $n$, and the last one is either $n-1$ or $n+1$. $~~~~~$ The case with $n-1$ collides with (1), and the case with $n+1$ collides with (2). Hence there is only the solution $(n,n,n,\ldots,n)$.

Only the trivial permutation $(n, n, n, \cdots, n)$ works. Obviously if the sum of the squares is equal to $n^3$, we are done by Cauchy. Now, suppose there exist $a_1, a_2, \cdots, a_n$ not equal such that $$a_1^2+a_2^2+\cdots+a_n^2 = n^3+1.$$The transformation $(a_1, a_2) \to (n, a_1+a_2-n)$ for unequal $a_1, a_2$ decreases the sum of squares by at least $2(n-a)(n-b) > 1$, which contradicts minimality.