Solutions to Problems; Chapter (^8 387)
Let
s(x) = C-ldf(+n)
= 2” + u,-lx”-l + unun4x”-2 +.. - + u;-2u1x + uyuo.
Then g(x) is a manic polynomial over 2 which has the property ascribed
to f (4.
We now modify g(x) to obtain a polynomial which is sure to have a
real zero which is nonrational. To this end, observe that g(x) - g(0) - x
has at most finitely many zeros, so that there exists a prime p for which
g(p) -g(O) -p is nonzero. Let h(x) = g(x) -g(O) -p. Then h(x) is a manic
polynomial over 2 of degree n which assumes rational values if and only if
x is rational. Since h(0) < 0 and h(x) has positive leading coefficient, h(x)
must have a positive zero. By hypothesis, this zero must be rational and
thus must be an integer dividing the prime p, and so must be p. But this
contradicts h(p) # 0. The result follows.
Remark. Compare this with Problem 5.4.23.
- Let u = u/(bc- u2), v = b/(cu - b2), w = c/(ub - c2). Then VW - u2 =
s/(bc - u2), wu - v2 = s/(cu - b2), uv - w2 = s/(ub - c2) where
’ =(u3b3 + b3c3 + c3u3 - 3u2b2c2)
(ub - c”)(cu - b2)(bc - u2) *Suppose, if possible, that s = 0. Then ub + bc + cu = 0 G- ub - c2 =
-c(u + b + c), cu - b2 = -b(u + b + c), bc - u2 = -u(u + b + c) whence
u = v = w = -l/(u + b + c) and so u + v + w is nonzero contrary to
hypothesis. Hence s # 0, and so
u/(bc - u”) + v/(cu - b2) + w/(ub - c”) = (3uvw - u3 - v3 - w”)/s= -(u + v + w)(u2 + v2 + w2 - uv - uw - vw)/s = 0as required.
- Let x = 1 + l/l + x = (2 + x)/(1 + x), xk+l = 1 + l/x, for k =
1,2,... , n - 1. By induction, it can be shown that
xk = (Fk+2 + Fk+lx)/(Fk+l + Fkx),where{Fk}={1,1,2,3,5,8 ,... } is the Fibonacci sequence. Thus, the equa-
tion to be solved is x, = x, i.e. F,,+2 + F,+lx = (F,+l + F,,x)x, whence
x = +iEygE.
- If w is a zero of f(t), th en so also are w2 + w +^1 and w2 - w +^1 =
(w - 1)2 + (w - 1) + 1. s ince 2w = (w2 + w + 1) - (w2 - w + l), it
follows that 212~1 5 lw2 + w + II+ I w2 - w +^11 with equality if and only if