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
y−17=2xx2+17−17=(2xx−17)(x−17)<21(x−17)
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.