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Solutions to Problems; Chapter 3 291

so that t > -15. Thus, -15 5 t 5 8. Checking possibiEties leads to the
following polynomials:

m t3 - mt2 - mt - (m2 + 1)

-77 t3 + 77t2 + 77t - 5930 = (t + 10)(t2 + 67t - 593)

-13 t3 + 13t2 + 13t - 170 = (t + lO)(@ + 3t - 17)

-7 t3 + 7t2 + 7t - 50 = (t - 2)(t2 + 9t + 25)

-2 t3 + 2t2 + 2t - 5 = (t - l)(@ + 3t + 5)

0 &-l=(t-l)(t2+t+l)

1 t3 - t2 - t - 2 = (t - 2)(t2 + t + 1)

8.8. Suppose, if possible, that m is an integer zero off. Let k be any positive
integer. Determine an integer c so that r = m - kc is one of (1,2,3,... , k}.
Then f(r) = f(m - kc) E f(m) = 0 (mod k). The result follows by a
contradiction argument.

8.9. For some polynomial g(t),

f(t) - 12 = (t - a)(t - b)(t - c)(t - d)g(t).

If f(k) = 25, then substituting k into this equation yields a representation
of 13 as the product of at least four distinct integers, an impossibility.

8.10. Since m + n + k = 0, 1 is a zero of the quadratic. The other zero
must be k/m = -1 - n/m.

8.11. We must have

(ad - bc)(x - p)” = (5d - b)x2 + (8d - 10b)x + (14d - 7b).

The discriminant of the right side must vanish, so that

(4d - 5b)2 = 7(5d - b)(2d - b),

which simplifies to (3d - 2b)(2d + b) = 0. Similarly,

(ad - bc)(x - q)” = (a - 5c)x2 + (lOa - 8c)z + (7a - 14~)

yields (3c - 2a)(2c + u) = 0.
Further information is obtained by comparing quadratic and constant
coefficients in the two equations:

a+b=5 c+d=l

up2 + bq2 = 14 cp2 + dq2 = 7.

Experimenting with IpI = 1, IqJ = 2 leads to the possibility

(a, h c, d,p, q) = (2,3, -1,2,1, -2).