1984 AIME Problem 15

Regard the four left sides as values of one rational function in $T$

Compare the coefficient of $\frac{1}{T}$ as $T$ tends to infinity

Define $$\begin{aligned} R(T)&=\frac{x^2}{T-1}+\frac{y^2}{T-9}\\ &\quad{}+\frac{z^2}{T-25}+\frac{w^2}{T-49}\\ &\quad{}-1. \end{aligned}$$ Put $$\begin{aligned} N(T)&=(T-4)(T-16)\\ &\quad{}\cdot(T-36)(T-64),\\ D(T)&=(T-1)(T-9)\\ &\quad{}\cdot(T-25)(T-49). \end{aligned}$$ With common denominator $D(T),$ the numerator of $R(T)$ has leading coefficient $-1.$ The four equations say its zeros are $4,$ $16,$ $36,$ and $64.$ Therefore $R(T)=-\frac{N(T)}{D(T)}.$

Let $S=w^2+x^2+y^2+z^2.$ As $T$ tends to infinity, the defining expression gives $$R(T)=-1+\frac{S}{T}+O(T^{-2}).$$ On the other hand, the sum of the four numerator roots is $120,$ while the sum of the four denominator roots is $84,$ so the factored expression gives $$\begin{aligned} R(T)&=-1+\frac{120-84}{T}\\ &\quad{}+O(T^{-2}). \end{aligned}$$ Hence $w^2+x^2+y^2+z^2=36.$