Find $\alpha , \beta$ s.t. $\forall s_i\in\mathbb{Z} ,\frac{\alpha^2}{\beta}\neq\frac{(s_1-s_2)^2+(s_3-s_4)^2+...}{(s_1+s_2)^2+(s_3+s_4)^2+...}$

Question

Let us assume that $\alpha,\beta , s_i\in\mathbb{Z}$ , for $i=1,...,8$. is it possible to choose $\alpha,\beta$ such that for all $s_i\in\mathbb{Z}$ the following equation is $never$ satisfied?$$\frac{\alpha^2}{\beta}=\frac{(s_1-s_2)^2+(s_3-s_4)^2+(s_5-s_6)^2+(s_7-s_8)^2}{(s_1+s_2)^2+(s_3+s_4)^2+(s_5+s_6)^2+(s_7+s_8)^2}$$ Note that $\alpha,\beta$ are $positive$ numbers.Assuming that the denominator is not zero.

Answer 1

First note that given $a,b\in\mathbb{Z}$ there is always a solution to $$ s_{1}-s_{2}=a $$ $$ s_{1}+s_{2}=b $$

Now, given $\alpha^{2},\beta\in\mathbb{Z}$ such that $\frac{\alpha^{2}}{\beta}$ is positive (and hence $\beta$ is positive) there are, by Lagrange's four-square theorem, integers such that $$ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=\alpha^{2} $$

$$ y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2}=\beta $$

Now, choosing $s_{1},s_{2}$ s.t $$ s_{1}-s_{2}=x_{1} $$ $$ s_{1}+s_{2}=y_{1} $$

and similarly for the others $s_{i}$. We see that there always $s_{i}$ that satisfies the above equation.

Moreover, this holds even if $\alpha\in\mathbb{Z}$ is positive and not just $\alpha^{2}\in\mathbb{Z}$