Each of the numbers x_1, x_2, . x_ 101 is ± 1 . What is the...

Let $S=\sum_{1 \leq i \leq j \leq 10} x_i x_j$ We have $$\begin{aligned} & \left(x_1+x_2+x_3+\ldots \ldots x_{101}\right)^2=x_1^2+x_2^2+x_3^2+\ldots \ldots . .+x_{101}^2+2 S \\ & \Rightarrow 2 S=\left(\sum_{i=1}^{101} x_i\right)^2-\sum_{i=1}^{101} x_i^2 \end{aligned}$$ Since we have $x_i= \pm 1$, So $x_i^2=1$ So $2 S=\left(\sum_{i=1}^{101} x_i\right)^2-101$ Since $\sum_{\mathrm{i}=1}^{101} \mathrm{x}_{\mathrm{i}}$ will be an integer So $\left(\sum_{i=1}^{101} x_i\right)^2$ will be a perfect square. For smallest positive $S:\left(\sum_{i=1}^{101} x_i\right)^2$ must be the smallest perfect square greater than 101 . So $\left(\sum_{i=1}^{101} x_i\right)^2=121 \Rightarrow \sum_{\mathrm{i}=1}^{101} \mathrm{x}_{\mathrm{i}}=11 \text { or }-11$ We can verify that the desired sum can be achieved by putting $45 x_i$ 's to be -1 and $56 x_i$ 's to be 1 So, 2S = 121-101=20 $\Rightarrow S=10