Oh, How Long It Takes?

I present two Solution over here:

Solution 1 :

Rewrite the system of equations as $\dfrac{x^{2}}{t-1}+\dfrac{y^{2}}{t-3^{2}}+\dfrac{z^{2}}{t-5^{2}}+\dfrac{w^{2}}{t-7^{2}}=1$ This equation is satisfied when $t = 4,16,36,64$ , as then the equation is equivalent to the given equations. After clearing fractions, for each of the values $t=4,16,36,64$ , we have the equation $x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)\$ \$+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)$ . When $t=4,16,36,64$ , a $(t-4)(t-16)(t-36)(t-64)$ term can be subtracted from the right-hand side because it equals 0. Thus we have the following equation which holds for $t=4,16,36,64$ :

$x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)\$ \$+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) \\ =(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)$

Each side of this equation is a polynomial in $t$ of degree at most 3, and they are equal for 4 values of $t$ (when $t=4,16,36,64$ ). Therefore, the polynomials must be equal for all $t$ .

Now we can plug in $t=1$ into the polynomial equation. Most terms drop, and we end up with

$x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)$

so that

$x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}$

Similarly, we can plug in $t=9,25,49$ and get

$\begin{aligned} y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\ z^2&=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\ w^2&=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{aligned}$

Now adding them up,

$\begin{aligned}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\ x^2+y^2&=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{aligned}$

with a sum of

$\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{36}.$

Lengthy proof that any two cubic polynomials in $t$ which are equal at 4 values of $t$ are themselves equivalent: Let the two polynomials be $A(t)$ and $B(t)$ and let them be equal at $t=a,b,c,d$ . Thus we have $A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0$ . Also the polynomial $A(t) - B(t)$ is cubic, but it equals 0 at 4 values of $t$ . Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into $(t-a)(t-b)(t-c)(t-d)$ (some nonzero polynomial) which would have a degree greater than or equal to 4, contradicting the statement that $A(t) - B(t)$ is cubic. Because $A(t) - B(t) = 0, A(t)$ and $B(t)$ are equivalent and must be equal for all $t$ .

Post script for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes $x^2,y^2,z^2$ and $w^2$ separately before adding them to obtain the final answer is appealing because it gives the individual values of $x^2,y^2,z^2$ and $w^2$ which can be plugged into the given equations to check.

Solution 2

As in Solution 1, we have

$(-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)\\=(t-4)(t-16)(t-36)(t-64)$

Now the coefficient of $t^3$ on both sides must be equal. Therefore we have $1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64$ implies $x^2+y^2+z^2+w^2=\boxed{36}$ .