Evaluating $a^2+b^2+c^2+d^2$ for Different Equations

[anemone]

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

$\dfrac{a^2}{2^2-1^2}+\dfrac{b^2}{2^2-3^2}+\dfrac{c^2}{2^2-5^2}+\dfrac{d^2}{2^2-7^2}=1$

$\dfrac{a^2}{4^2-1^2}+\dfrac{b^2}{4^2-3^2}+\dfrac{c^2}{4^2-5^2}+\dfrac{d^2}{4^2-7^2}=1$

$\dfrac{a^2}{6^2-1^2}+\dfrac{b^2}{6^2-3^2}+\dfrac{c^2}{6^2-5^2}+\dfrac{d^2}{6^2-7^2}=1$

$\dfrac{a^2}{8^2-1^2}+\dfrac{b^2}{8^2-3^2}+\dfrac{c^2}{8^2-5^2}+\dfrac{d^2}{8^2-7^2}=1$

 

[Opalg]

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

$\dfrac{a^2}{2^2-1^2}+\dfrac{b^2}{2^2-3^2}+\dfrac{c^2}{2^2-5^2}+\dfrac{d^2}{2^2-7^2}=1$

$\dfrac{a^2}{4^2-1^2}+\dfrac{b^2}{4^2-3^2}+\dfrac{c^2}{4^2-5^2}+\dfrac{d^2}{4^2-7^2}=1$

$\dfrac{a^2}{6^2-1^2}+\dfrac{b^2}{6^2-3^2}+\dfrac{c^2}{6^2-5^2}+\dfrac{d^2}{6^2-7^2}=1$

$\dfrac{a^2}{8^2-1^2}+\dfrac{b^2}{8^2-3^2}+\dfrac{c^2}{8^2-5^2}+\dfrac{d^2}{8^2-7^2}=1$

[sp]Let $f(x) = \dfrac{a^2}{x-1^2} + \dfrac{b^2}{x-3^2} + \dfrac{c^2}{x-5^2} + \dfrac{d^2}{x-7^2}$. Then $f(x) = 1$ when $x = 2^2, 4^2,6^2$ or $8^2$.

Multiplying through by $(x-1^2)(x-3^2)(x-5^2)(x-7^2)$, the equation $f(x) = 1$ becomes $$\begin{aligned} a^2(x-3^2)(x-5^2)(x-7^2) &+ b^2(x-1^2)(x-5^2)(x-7^2) + c^2(x-1^2)(x-3^2)(x-7^2) + d^2(x-1^2)(x-3^2)(x-5^2) \\&= (x-1^2)(x-3^2)(x-5^2)(x-7^2).\end{aligned}$$ That is a quartic equation, whose first two terms are $$x^4 - (a^2 + b^2 + c^2 + d^2 + 1^2 + 3^2 + 5^2 + 7^2)x^3 + \ldots = 0.$$ The sum of the roots of that equation is $a^2 + b^2 + c^2 + d^2 + 1^2 + 3^2 + 5^2 + 7^2.$ But the four roots are $2^2, 4^2,6^2$ and $8^2$. Therefore $$a^2 + b^2 + c^2 + d^2 + (1^2 + 3^2 + 5^2 + 7^2) = 2^2 + 4^2 + 6^2 + 8^2,$$ and so $$a^2 + b^2 + c^2 + d^2 = (2^2 - 1^2) + (4^2 - 3^2) + (6^2 - 5^2) + (8^2 - 7^2) = 3 + 7 + 11 + 15 = 36.$$[/sp]

 

[anemone]

Very well done, Opalg and thanks for participating! (Smile)