What is the Sum of Squares in POTW #267?

[anemone]

Here is this week's POTW:

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Determine $a^2+b^2+c^2+d^2$ if

\(\displaystyle \dfrac{a^2}{2^2-1^2}+\dfrac{b^2}{2^2-3^2}+\dfrac{c^2}{2^2-5^2}+\dfrac{d^2}{2^2-7^2}=1\\\dfrac{a^2}{4^2-1^2}+\dfrac{b^2}{4^2-3^2}+\dfrac{c^2}{4^2-5^2}+\dfrac{d^2}{4^2-7^2}=1\\\dfrac{a^2}{6^2-1^2}+\dfrac{b^2}{6^2-3^2}+\dfrac{c^2}{6^2-5^2}+\dfrac{d^2}{6^2-7^2}=1\\\dfrac{a^2}{8^2-1^2}+\dfrac{b^2}{8^2-3^2}+\dfrac{c^2}{8^2-5^2}+\dfrac{d^2}{8^2-7^2}=1\)

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[anemone]

No one answered last week's problem.(Sadface)

You can find the suggested solution below:

Spoiler

The claim that the given system of equations is satisfied by $a^2,\,b^2,\,c^2$ and $d^2$ is equivalent to claiming that

\(\displaystyle \dfrac{a^2}{t-1^2}+\dfrac{b^2}{t-3^2}+\dfrac{c^2}{t-5^2}+\dfrac{d^2}{t-7^2}=1 \tag{1}\)

is satisfied by $t=4,\, 16,\,36$ and $64$.

Clearing the fractions, we find that for all values of $t$ for which it is defined (i.e. $t\ne 1,\,9,\,25$ and $49$), $(1)$ is equivalent to the polynomial equation $P(t)=0$, where

$P(t)=(t-1)(t-9)(t-25)(t-49)-a^2(t-9)(t-25)(t-49)-b^2(t-1)(t-25)(t-49)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-c^2(t-1)(t-9)(t-49)-d^2(t-1)(t-9)(t-25)$

Since degree $P(t)=4,\,P(t)=0$ has exactly four zeros $t=4,\,16,\,36$ and $64$, i.e.,

$P(t)=(t-4)(t-16)(t-36)(t-64)$

Comparing the coefficients of $t^3$ in the two expressions of $P(t)$ yields

$1+9+25+49+a^2+b^2+c^2+d^2=4+16+36+64$,

from which it follows that

$a^2+b^2+c^2+d^2=36$