# Problem Of The Week #423 June 29th, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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Determine $x^2+y^2+z^2+w^2$ if

$\dfrac{x^2}{2^2-1^2}+\dfrac{y^2}{2^2-3^2}+\dfrac{z^2}{2^2-5^2}+\dfrac{w^2}{2^2-7^2}=1,\\\dfrac{x^2}{4^2-1^2}+\dfrac{y^2}{4^2-3^2}+\dfrac{z^2}{4^2-5^2}+\dfrac{w^2}{4^2-7^2}=1,\\\dfrac{x^2}{6^2-1^2}+\dfrac{y^2}{6^2-3^2}+\dfrac{z^2}{6^2-5^2}+\dfrac{w^2}{6^2-7^2}=1,\\\dfrac{x^2}{8^2-1^2}+\dfrac{y^2}{8^2-3^2}+\dfrac{z^2}{8^2-5^2}+\dfrac{w^2}{8^2-7^2}=1$

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#### anemone

##### MHB POTW Director
Staff member
No one answered last week's problem.

You can find the suggested solution below:

The claim that the given system of equations is satisfied by $x^2,\,y^2,\,z^2$ and $w^2$ is equivalent to claiming that

$$\displaystyle \dfrac{x^2}{t-1^2}+\dfrac{y^2}{t-3^2}+\dfrac{z^2}{t-5^2}+\dfrac{w^2}{t-7^2}=1 \tag{1}$$

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

$P(t)=(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)$

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

$1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64$,

from which it follows that

$x^2+y^2+z^2+w^2=36$

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