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 Dependent Variable

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# Polynomials

1 Warm-Ups

1. Consider the cubic equation ax3 + bx2 + cx + d = 0. The roots are Prove that no such general formula exists for a quintic equation.

2 Theory

Thanks to Elgin Johnston (1997) for these theorems.

Rational Root Theorem Let be a polynomial with integer coefficients.
Then any rational solution r/s (expressed in lowest terms) must have and .

Descartes’s Rule of Signs Let be a polynomial with real coefficients.
Then the number of positive roots is equal to N − 2k, where N is the number of sign changes in the
coefficient list (ignoring zeros), and k is some nonnegative integer.

Eisenstein’s Irreducibility Criterion Let be a polynomial with integer
coefficients and let q be a prime. If q is a factor of each of but q is not a factor of
an, and q2 is not a factor of a0, then p(x) is irreducible over the rationals.

Einstein’s Theory of Relativity Unfortunately, this topic is beyond the scope of this program.

Gauss’s Theorem If p(x) has integer coefficients and p(x) can be factored over the rationals, then p(x)
can be factored over the integers.

Lagrange Interpolation Suppose we want a degree-n polynomial that passes through a set of n+1 points: . Then the polynomial is: where the i-th “normalization” factor is the product of all the terms that have j ≠ i.

3 Problems

Thanks to Elgin Johnston (1997) for most of these problems.

1. (Crux Math., June/July 1978) Show that is composite when n is any integer.
Solution: Factor as difference of two squares. Prove that neither factor can be ±1.

2. (St. Petersburg City Math Olympiad 1998/14) Find all polynomials P(x, y) in two variables such that
for any x and y, P(x + y, y − x) = P(x, y).

Solution: Clearly constant polynomials work. Also, P(x, y) = P(x + y, y − x) = P(2y,−2x) =
P(16x, 16y). Suppose we have a nontrivial polynomial. Then on the unit circle, it is bounded because
we can just look at the fixed coefficients. Yet along each ray y = tx, we get a polynomial whose translate
has infinitely many zeros, so it must be constant. Hence P is constant along all rays, implying that P
is bounded by its max on the unit circle, hence bounded everywhere. Now suppose maximum degree
of y is N. Study the polynomial . The leading coeff of this is equal to the leading coeff of
P(x, y) when sorted with respect to x as more important. Since the z-poly is also bounded everywhere,
it too must be constant, implying that the leading term is a constant.

3. (Putnam, May 1977) Determine all solutions of the system Solution: Given solutions x, y, z, construct 3-degree polynomial P(t) = (t − x)(t − y)(t − z). Then In particular, roots are w and a pair of opposites.

4. (Crux Math., April 1979) Determine the triples of integers (x, y, z) satisfying the equation Solution: Move z3 to RHS and factor as x3 ±y3. We get (x+y) = 0 or (y +z)(z +x) = 0. So two
are opposites.

5. (USSR Olympiad) Prove that the fraction (n3+2n)/(n4+3n2+1) is in lowest terms for every positive
integer n.

Solution: Use Euclidean algorithm for GCD. (n3 + 2n)n = n4 + 2n2, so difference to denominator
is n2 + 1. Yet that’s relatively prime to n(n2 + 2).

6. (Po, 2004) Prove that is irreducible.
Solution: Eisenstein with substitution x → x + 1.

7. (Canadian Olympiad, 1970) Let P(x) be a polynomial with integral coefficients. Suppose there exist
four distinct integers a, b, c, d with P(a) = P(b) = P(c) = P(d) = 5. Prove that there is no integer k
with P(k) = 8.

Solution: Drop it down to 4 zeros, and check whether one value can be 3. Factor as P(x) =
(x − a)(x − b)(x − c)(x − d)R(x); then substitute k. 3 is prime, but we’ll get at most two ±1 terms
from the (x − α) product.

8. (Monthly, October 1962) Prove that every polynomial over the complex numbers has a nonzero polynomial
multiple whose exponents are all divisible by 109.

Solution: Factor polynomial as . Then the desired polynomial is , where P = 109. Each factor divides the corresponding factor.

9. (Elgin, MOP 1997) For which n is the polynomial divisible by the polynomial ?

Solution: Observe: So if the quotient is Q(x), then Q(x)(x + 1) = xn + 1. This happens iff .1 is a root of xn + 1, which
is iff n is odd.

10. (Czech-Slovak Match, 1998/1) A polynomial P(x) of degree n ≥5 with integer coefficients and n
distinct integer roots is given. Find all integer roots of P(P(x)) given that 0 is a root of P(x).

Solution: Answer: just the roots of P(x). Proof: write . Suppose
we have another integer root r; then for some k. Since degree is at least 5,
this means that we have dividing rk. Simple analysis shows that r is between 0 and rk; more
analysis shows that we just need to defuse the case of 2ab | a + b. Assume a ≤b. Now if a = 1, only
solution is b = 1, but then we already used ±1 in the factors, so we actually have to have dividing rk, no good. If a > 1, then 2ab > 2b ≥a + b, contradiction.

11. (Hungarian Olympiad, 1899) Let r and s be the roots of Prove that r3 and s3 are the roots of Hint: use Linear Algebra.
Solution: r and s are the eigenvalues of the matrix The y equation is the characteristic polynomial of the cube of that matrix.

12. (Hungarian Olympiad, 1981) Show that there is only one natural number n such that is
a perfect square.

Solution: . So, need to have 2n as difference of squares . Hence (N + 48),
(N − 48) are both powers of 2. Their difference is 96. Difference between two powers of 2 is of the
form . Uniquely set to .

13. (MOP 97/9/3) Let be a set of n distinct complex numbers, for some n ≥9, exactly
n − 3 of which are real. Prove that there are at most two quadratic polynomials f(z) with complex
coefficients such that f(S) = S (that is, f permutes the elements of S).

14. (MOP 97/9/1) Let be a nonzero polynomial with integer coefficients
such that P(r) = P(s) = 0 for some integers r and s, with 0 < r < s. Prove that for some k.