Solutions to Problems; Chapter (^8 383)
slightly exceeds any one of the integers 1,2,... ,70 and not satisfied when
z is slightly less than one of the integers 2,3,... ,70. Let
p(x)=5(x-1)...(I-h)...(r-70)-4~h(z-1)...(x~~)...(1-7o).
k=l
Thenp(r)>Owhenr=1,3,5 ,..., 69andp(r)<Owhenr=2,4,6 ,..., 70.
Since degp(x) = 70 and p(z) > 0 for large 2, p(x) has exactly one real zero
u,. in each of the open intervals (r, r + 1) (1 < r 5 69) and a real zero
urc > 70. The inequality will be satisfied exactly when x belongs to one of
the intervals (r, u,] (1 5 r 5 70). Let S = 1+2+. ..+70 = (35)(71). Since
the coefficient of x6ā in p(z) is -5s - 4s = -9S, we find that Cu, = 9S/5.
Hence the total length of all the intervals is
E(u,. - r) = 9S/5 - S = 4(7)(71) = 1988.
r=l
- First solution. Suppose a + b # 0. We have that
0 = (CU)~ = u3 + 3Cu2b + 6Cubc
=^0 + 3Ca2(-a) + 6Cubc = GCubc,where the summations are symmetric in the variables a, b, c, d, e, f. The
left side multiplied by a + b is(Q + b)(u + C)(Q + d)(u + e)(u + f)= u5 + u4(Cb) + u3(Cbc) + u2(Cbcd) + u(Cbcde) + bcdef
= (u5 - uā) + u2(uCbc + bed) + (dbcde + bcdef)
= 0 + 0 + (uCbcde + bcdef),where the summations are symmetric in the variables b, c, d, e, f. By
symmetry, the right side multiplied by a + b is the same, and the result
follows. If a + b = 0, then the given conditions are c + d + e + f = c3 +
d3 + e3 + f3 = 0 and we find that the coefficients of even powers of (o + c)
(o + d)(a + e)(u + f) and (e - ~)(a - d)(u - e)(a - f) agree while those of
odd powers vanish by a similar argument.
Second solution. Noting that (u + c + d) + (b + e + f) is a factor of the
sum of the corresponding cubes, we have that
0 = (u + c + d)3 + (b + e + f)3
= 3[(u + C)(U + d)(c + d) + (b + e)(b + f>(e + f)]whence
(u + c)(u + d)(c + d) = -(b + e)(b + f)(e + f).