Problem:
Determine x2+y2+z2+w2 if
​22−12x2​+22−32y2​+22−52z2​+22−72w2​=1,42−12x2​+42−32y2​+42−52z2​+42−72w2​=1,62−12x2​+62−32y2​+62−52z2​+62−72w2​=1,82−12x2​+82−32y2​+82−52z2​+82−72w2​=1.​
Solution:
The claim that the given system of equations is satisfied by x2,y2,z2 and w2 is equivalent to claiming that
t−1x2​+t−9y2​+t−25z2​+t−49w2​=1(1)
is satisfied by t=4,16,36 and 64. Multiplying to clear fractions,
we find that for all values of t for which it is defined (i.e., tî€ =1,9,25 and 49), Equation (1) is equivalent to the polynomial equation
​(t−1)(t−9)(t−25)(t−49)−x2(t−9)(t−25)(t−49)−y2(t−1)(t−25)(t−49)−z2(t−1)(t−9)(t−49)−w2(t−1)(t−9)(t−25)=0​(2)
where the left member may be viewed as a fourth degree polynomial in t. Since t=4,16,36 and 64 are known to be roots, and a 4th degree polynomial can have at most 4 roots, these must be all the roots. It follows that (2) is equivalent to
(t−4)(t−16)(t−36)(t−64)=0(3)
Since the coefficient of t4 is 1 in both (2) and (3), one may conclude that the coefficients of the other powers of t must also be the same. In particular, equating the coefficients of t3, we have
1+9+25+49+x2+y2+z2+w2=4+16+36+64
from which x2+y2+z2+w2=36​.
Note. By equating the left members of Equations (2) and (3) and letting t=1, one also finds that x2=102411025​. Similarly, letting t=9,25 and 49 in succession yields y2=102410395​,z2=10249009​ and w2=10246435​. One can show that these values indeed satisfy the given system of equations, and that their sum is 36.
The problems on this page are the property of the MAA's American Mathematics Competitions