Find $a^2 + b^2 + c^2 + d^2$. 
$$\begin{align}\dfrac{a^2}{2^2 - 1^2} + \dfrac{b^2}{4^2 - 1^2} + \dfrac{c^2}{6^2 - 1^2} + \dfrac{d^2}{8^2 - 1^2} &= \dfrac{a^2}{2^2 - 3^2} + \dfrac{b^2}{4^2 - 3^2} + \dfrac{c^2}{6^2 - 3^2} + \dfrac{d^2}{8^2 - 3^2}
\\&= \dfrac{a^2}{2^2 - 5^2} + \dfrac{b^2}{4^2 - 5^2} + \dfrac{c^2}{6^2 - 5^2} + \dfrac{d^2}{8^2 - 5^2} \\ &= \dfrac{a^2}{2^2 - 7^2} + \dfrac{b^2}{4^2 - 7^2} + \dfrac{c^2}{6^2 - 7^2} + \dfrac{d^2}{8^2 - 7^2}
\\ &= 1\end{align}$$
Find $a^2 + b^2 + c^2 + d^2$.

My Approach:
I tried forming Linear equations by taking $a^2=x, b^2=y, c^2=z, d^2=w$
Then I obtained
$\dfrac{x}{3}+\dfrac{y}{15}+\dfrac{z}{35}+\dfrac{w}{63}=1$
$\dfrac{x}{-5}+\dfrac{y}{7}+\dfrac{z}{27}+\dfrac{w}{55}=1$
$\dfrac{x}{-21}+\dfrac{y}{-9}+\dfrac{z}{11}+\dfrac{w}{39}=1$
$\dfrac{x}{-45}+\dfrac{y}{-33}+\dfrac{z}{-13}+\dfrac{w}{15}=1$
After solving with www.wolframalpha.com I obtained the result
$w=\frac{6435}{256}$, $x=\frac{315}{512}$, $y= \frac{693}{256}$, $z=\frac{3861}{512}$
Which lead me $x+y+z+w=36$ that is $a^2+b^2+c^2+z^2=36$
But how to solve this question without any calculator or online tools because it was asked to a $11th$ class student
Wolfram Alpha

We write down the following polynomial:
\begin{eqnarray}
f(x) &=& (x - 2^2)(x - 4^2)(x - 6^2)(x - 8^2)\\
&+& a^2(x - 4^2)(x - 6^2)(x - 8^2)\\
&+& b^2(x - 2^2)(x - 6^2)(x - 8^2)\\
&+& c^2(x - 2^2)(x - 4^2)(x - 8^2)\\
&+& d^2(x - 2^2)(x - 4^2)(x - 6^2).
\end{eqnarray}
The given conditions then imply that $f(1^2) = f(3^2) = f(5^2) = f(7^2) = 0$.
However, $f(x)$ is a monic polynomial of degree $4$. Therefore we have $$f(x) = (x - 1^2)(x - 3^2)(x - 5^2)(x - 7^2).$$
It only remains to evaluate $f(x)$ at $x = 2^2, 4^2, 6^2, 8^2$ respectively to get values of $a^2, b^2, c^2, d^2$.

As suggested in the comments, one can look at the coefficient before $x^3$ to get the value of $a^2 + b^2 + c^2 + d^2$ without doing much calculation.
