 
Summary: SOLUTION TO PROBLEM 1546
PROPOSED BY B. G. KLEIN, ET AL
Tewodros Amdeberhan
DeVry Institute, Mathematics
630 US Highway One, North Brunswick, NJ 08902
amdberhan@admin.nj.devry.edu
Problem 1546: P Given y 1, let P be the set of all real polynomials px with nonnegative
coe cients that satisfy p1 = 1 and p3 = y. Prove there exists p0x 2 P such that
i fp2 : px 2 Pg = 1;p02 ;
ii if px 2 P and p2 = p02, then px = p0x.
Solution: We show that p0x = y,1
2 x , 1 + 1 is the unique polynomial satisfying the above
conditions with p02 = y + 1=2.
Note that since coe cients are nonnegative, all functions in P are strictly increasing as well as
concave upwards for 1 x 3. Thus certainly for each px 2 P, we have px p0x;x 2 1;3
since the line p0x joins the end points 1;1 and 3;y.
Furthermore, p0x is unique. Else assume that p2 = p02, for some p0x 6= px 2 P. Then
by the IntermediateValueTheorem for derivatives, there exist two distinct points one in 1;2 and
another in 2;3 where tangents to px hence derivatives have same slope, that is, y , 1=2.
This cannot be true of the nonlinear px as its derivative is onetoone because p00x 0 there.
