Fractional Part, Integrals, and Number Theory

One fun way to think about this integral is to look at $I_k = \int_{k-1}^{k} x\{x\}\,dx,$ where $k\in \{1,2,\ldots,N\}.$ Informally speaking, we're integrating as $\{x\}$ goes from $0\rightarrow 1,$ each time adding 1 to the multiplier value as we go along (as $k$ goes from $1\rightarrow N$ ), so each piece of the integral $I_{k+1}$ will be $\int_{0}^{1}x\,dx = \frac{1}{2}$ more than the previous piece $I_{k}.$

More formally, $\begin{aligned} I_{k+1}-I_k &= \int_{k}^{k+1} x\{x\}\,dx - \int_{k-1}^{k} x\{x\}\,dx \\ &= \underbrace{\int_{k-1}^{k}(x+1)\{x\}\,dx}_{x\rightarrow x+1\ , \ \{x+1\} = \{x\}} - \int_{k-1}^{k}x\{x\}\, dx \\ &= \int_{k-1}^{k} \{x\}\,dx = \int_{0}^{1} x \,dx = \frac{1}{2}. \end{aligned}$

Now, note that $I_1 = \int_0^1 x\{x\} \,dx = \int_0^1 x^2 \,dx = \frac{1}{3}.$ Then, our integral is $I_1 + I_2 + \cdots + I_N = \frac{1}{3} + \left(\frac{1}{3}+ \frac{1}{2}\right) + \cdots + \left(\frac{1}{3} + (N-1)\cdot \frac{1}{2}\right) = \frac{N}{3} + \frac{(N-1)(N)}{4} = \frac{3N^2+N}{12}.$

Since $12 = 3\cdot 4$ , we need $3N^2 + N$ to be divisible by $3$ and $4.$ Since $3|3N^2,$ we must have $3|N.$ It's quick to check $N=3,6,9,\ldots$ and see that $N=9$ is the smallest solution.

On the other hand, if we had a perfect square $\frac{3N^2+N}{12}=M^2,$ this would imply that $3N^2 + N = 12M^2 \ \iff \ N = 3\left(4M^2 - N^2\right) = 3 \left(2M+N\right)\left(2M-N\right).$ However, this is impossible, since $N$ cannot be the product of three integers where one of the integers is greater than $N$ (here, $2M+N> N.$ )

Thus, $X+Y = 9+0 = 9.$