**1 Warm-Ups**

1. Consider the cubic equation ax^{3} + bx^{2} + 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

a_{n}, and q^{2} is not a factor of a_{0}, 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 z^{3} to RHS and factor as x^{3} ±y^{3}. We get (x+y) = 0 or (y +z)(z +x)
= 0. So two

are opposites.

5. (USSR Olympiad) Prove that the fraction (n^{3}+2n)/(n^{4}+3n^{2}+1) is in lowest terms
for every positive

integer n.

**Solution:** Use Euclidean algorithm for GCD. (n^{3} + 2n)n = n^{4} + 2n^{2}, so difference
to denominator

is n^{2} + 1. Yet that’s relatively prime to n(n^{2} + 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 10^{9}.

**Solution:** Factor polynomial as
. Then the
desired polynomial is

, where P = 10^{9}. 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) = x^{n } + 1.
This happens iff .1 is a root of x^{n} + 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 r_{k}. Simple analysis shows that r is
between 0 and r_{k}; 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 r_{k}, 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 r^{3} and s^{3} 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 2^{n} 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.