Deep Fact: 1 + 2 + ... + 9 = 45

A deeper fact: 45=45. Seriously, I don't understand this enigma!!

kingu wrote: Deep Fact: 1 + 2 + ... + 9 = 45 Zero is not a positive integer (at least not in the context of this problem). Perhaps a better problem statement would be: New problem statement wrote: Find the smallest $n\in\mathbb{N}$, for which there exist pairwise distinct positive integers $a_1, a_2, \ldots, a_n$ such that $$\frac{(a_1+a_2+\dots+a_n)^2-2025}{a_1^2+a_2^2+\dots +a_n^2 } \in\mathbb{N}. $$

Marinchoo wrote: Perhaps a better problem statement would be: New problem statement wrote: Find the smallest $n\in\mathbb{N}$, for which there exist pairwise distinct positive integers $a_1, a_2, \ldots, a_n$ such that $$\frac{(a_1+a_2+\dots+a_n)^2-2025}{a_1^2+a_2^2+\dots +a_n^2 } \in\mathbb{N}. $$ That gives more ambiguity I think

$$ \begin{aligned} & \text{Let } \left( \sum a_i \right)^2 - 2025 = k \sum a_i^2, \quad k \in \mathbb{Z}^+. \\ & \sum a_i \equiv \sum a_i^2 \pmod{2} \Rightarrow \sum a_i \equiv \sum a_i^2 \equiv 1 \pmod{2}. \\ & \Rightarrow 8 \mid k \sum a_i^2 \Rightarrow 8 \mid k \Rightarrow k \geq 8 \Rightarrow n \geq 9. \\ & \text{Then } \left( a_1, a_2, \ldots, a_9 \right) = (15, 16, 17, 18, 19, 20, 21, 26, 31) \text{ works}. \end{aligned} $$

Alternative construction: we take $\{a_i\}=\{n,n\pm a,n\pm b,n\pm c,n\pm d\}$ with $0<a<b<c<d<n$. Then the condition $\textstyle(\sum a_i)^2-2025=8\sum a_i^2$ simplifies to $2025+16(a^2+b^2+c^2+d^2)=9n^2$, which is easy to find solutions for (eg. taking mod 16, we get $n\equiv\pm1\pmod8$); the smallest is $(n,a,b,c,d)=(23,3,4,5,11)$, which corresponds to $\{a_i\}=\{12,18,19,20,23,26,27,28,34\}$.

Write $\left(\sum_{i=1}^n a_i\right)^2 - 2025 = k\sum_{i=1}^n a_i^2$. Observe firstly that $\sum_{i=1}^n a_i$ must be odd - indeed, if it is even, then from $x^2 \equiv x \pmod 2$ we get that $\sum_{i=1}^n a_i^2$ is also even, so the left-hand side of the main equation is odd, while the right-hand one is even, contradiction. Hence both $\sum_{i=1}^n a_i$ and $\sum_{i=1}^n a_i^2$ are odd. Since $t^2 \equiv 1 \pmod 8$ for odd $t$, the left-hand side is divisible by $8$, so from the right-hand side we obtain that $8$ divides $k$. In particular, $k\geq 8$. However, $\sum_{i=1}^n a_i^2 \geq \frac{\left(\sum_{i=1}^n a_i\right)^2}{n}$ from the QM-AM inequality, so with $S = \sum_{i=1}^n a_i$ we obtain as a necessary condition $S^2 > S^2 - 2025 \geq \frac{8S^2}{n}$, i.e. $n\geq 9$. (Also, for equality with $n=9$, the value $k=8$ is necessary.) The simplest equality case with $n=9$ is $x_1 = x_2 = \ldots = x_9 = 15$. A slightly more complicated example is $x_1 = x_2 = \ldots = x_8 = 4 \cdot 15 = 60$ and $x_9 = 9 \cdot 15 = 135$. If one really wants to find an example with all numbers distinct (as the problem requests), one way is to search for $(x - 1, x , x + 1, y - 1, y, y + 1, z-1, z, z + 1)$, then manipulations and extracting perfect squares lead to $x = 16$, $y = 21$, $z = 30$, i.e. $(15, 16, 17, 20, 21, 22, 29, 30, 31)$.

Great problem! Unfortunately, only a single $11$th grader managed to solve it on the actual competition.