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May 18th

May 18th

# Math Problem Set 8

1. (a) Factor the polynomial x 8 +98x4 +1 into two factors with integer (not necessarily
real) coefficients .
(b) Find the remainder on dividing x 100 − 2x51 + 1 by x2 − 1. (Shelley)

Solution:

What is the remainder?
x2 − 1 = (x + 1)(x − 1). So two solutions are x = ±1. Using Bezout’s theorem, we
substitute these into the original function (the solution must be of the form bx + c
because it must be of a lesser degree than the divisor):

f(1) = 0 = bx + c = b + c
f(−1) = 4 = −bx + c = −b + c

Solving the system of equations , b= -2, c=2. The remainder is -2x + 2.

2. If x1 and x2 are the zeros of the polynomial x2 − 6x + 1, then for every non negative
integer n, is an integer and not divisible by 5. (Derek) (Hint: how about
induction?)

3. (VA 1982) Let p(x) be a polynomial of the form p(x) = ax2 + bx + c, where a, b and
c are integers, with the property that 1 < p(1) < p(p(1)) < p(p(p(1))). Show that
a ≥ 0. (Brett)

We don’t know how to do this one yet !

4. (VA 1987) A sequence of polynomials is given by for
n ≥ 0, where a0 = a1 = 1 and, for n ≥ 0, . Denote by rn and sn the
roots of p n(x) = 0, with rn ≤ sn. Find (Ben)

Solution: Since this is a quadratic equation , we can make use of the quadratic formula
to come up with equations for the two roots as follows:

where we have used the fact that
Thus we have,

and

And for the other root we have

Now, we need a formula for an, and we know that

(0.1)

The solution to equation (1) is given by where and are roots
of the characteristic equation , and c1 and c2 are to be de termined
by initial conditions (although we will not need them for this problem). So solving
the characteristic for its roots via the quadratic formula, we find that:

Thus,

(0.2)

Notice that in equation (2) it is the case that and since we are evaluating
we can neglect the term.

Thus we have

5. (VA 1991) Prove that if is a real root of (1−x2)(1+x+x2+· · ·+xn)−x = 0 which lies
in (0, 1), with n = 1, 2, · · · , then is also a root of (1−x2)(1+x+x2+· · ·+xn+1)−1 = 0.
(Lei) (Hint: use 1 + x + x2 + · · · + xn = (1 − xn+1)/(1 − x).)

6. (VA 1996) Let ai, i = 1, 2, 3, 4, be real numbers such that
Show that for arbitrary real numbers bi, i = 1, 2, 3, the equation
has at least one real root which is on the interval
−1≤ x ≤ 1. (Tina)

Solution:

Taking the integral of the given equation over the interval −1≤ x ≤ 1 gives:

And by the Mean Value Theorem , since the integral of the equation is zero, the
function must take on the value of zero somewhere on the interval −1 ≤ x ≤ 1.

7. (VA 1995) Let Show that for every positive integer
n. Here [r] denotes the largest integer that is not larger than r. (David Rose)

Solution

8. Solve the equation z8 + 4z6 − 10z4 + 4z2 + 1 = 0. (Lei)

9. (Putnam 2004-B1) Let be a polynomial with integer
coefficients. Suppose that r is a rational number such that P(r) = 0. Show that the
n numbers

are integers. (David Edmonson)

Solution. Set i ∈ {0, 1, . . . , n - 1}. Now write r as u/v , where u and v are coprime.
Then
0, so
is a multiple of is also a
multiple of since and are coprime. Thus,
is an integer, which is the claim that we were asked to show.

10. (Putnam 2003-B1) Do there exist polynomials a(x), b(x), c(y), d(y) such that

1 + xy + x2y2 = a(x)c(y) + b(x)d(y)

holds identically? (Richard)

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