Problem:
The number of real solutions (x,y,z,w) of the simultaneous equations
2y=x+x17​,2z=y+y17​,2w=z+z17​,2x=w+w17​
is
Answer Choices:
A. 1
B. 2
C. 4
D. 8
E. 16
Solution:
Either x>0 or x<0. Also, for any positive number a, 21​(a+a17​)≥17​, with equality only if a=17​, because this inequality is equivalent to (a−17​)2≥0. Thus, if x>0, then considering each of the given equations in turn, one deduces that y≥17​,z≥17​,w≥17​ and x≥17​. Suppose x>17​. Then
so that x>y. Similarly, y>z,z>w, and w>x, implying x>x, an obvious contradiction. Therefore x=y=z=w=17​, clearly a solution, is the only solution with x>0. As for x<0, note that (x,y,z,w) is a solution if and only if (−x,−y,−z,−w) is a solution. Thus x=y=z=w=−17​ is the only other solution.