Problems posed at the St Andrews Colloquium, 1996

1. Somebody incorrectly remembered Fermat's little theorem as saying that the congruence aⁿ⁺¹= a (mod n) holds for all a if n is prime. Describe the set of integers n for which this property is in fact true.
Solution

2. Prove or disprove: for every odd number k there is a prime of the form 2ⁿ k+1.
Solution

3. Show that if (a² + b²)/(ab+1) (for a,b in N) is integral, then it is a perfect square.
Solution

1. Show that there is no ring (= commutative, with 1) with exactly five units.
Solution

2. Show that if a group has only finitely many elements of finite order, then these form a subgroup.
Solution

3. Let P and S denote the direct product and direct sum of countably many copies of Z, respectively, i.e. P is the set of all sequences of integers (a₁, a₂, a₃, ... ) with the abelian group structure given by componentwise addition, and S is the subgroup consisting of all sequences with a_i = 0 for i sufficiently large. Show that Hom_Z(P, Z) is isomorphic to S.
Solution

1. Associate to a prime the polynomial whose coefficients are the decimal digits of the prime (e.g. 9 x³ + 4 x² + 3 for the prime 9403). Show that this polynomial is always irreducible.
Solution

2. Suppose that b_n xⁿ + b_n-1 x^n-1 + ... + b₁ x + b₀ in R[x] has only real roots. Show that the polynomial b_n/n! xⁿ + b_n-1/(n-1)! x^n-1 + ... + b₁/1! x + b₀, has the same property.
Solution

3. Let p = 2m - 1 be an odd prime. Show that the polynomial (1-x) ^m + 1 + x^m is twice a square in F_p[x].
Solution

1. Let A and B be two adjacent vertices of an equilateral polygon. If the angles at all other vertices are known to be rational (when measured in degrees), show that the angles at A and B are also rational. Give a counterexample to this statement when A and B are not assumed to be adjacent.
Solution

2. A rectangle R is the union of finitely many smaller rectangles (non-overlapping except on their boundaries), each one of which has at least one rational side. Show that R has the same property.
Solution

3. Mark an angle a (0 < a < 2 ) on a pie-plate, and pick another angle b (0 < b < 2 ).
Define an operation on the pie as follows: cut out the slice of pie over the marked angle, lift it up, turn it over, replace it, and rotate the whole pie on the plate by the angle b. Show that, whatever the values of a and b, this operation has finite order (i.e., after a finite number of iterations every piece of the pie is in its original position).
Solution

1. Define a sequence t₁, t₂, t₃, ... by the recursion
t_n+5 = (t_n+4 t_n+1 + t_n+3 t_n+2)/t_n
with initial values (1,1,1,1,1). Prove the statement that all of the t_n are integral.
Solution

2. Define a sequence u₁, u₂, u₃, ... by the formula

Show that the statement "none of the u_n are integral" is false, but that the first counterexample is approximately 10^2019025.
Solution

3. Define a sequence v₁, v₂, v₃, ... by the recursion
v_n =(2+v₁² + ... + v_n-1²)/n.

Show that the statement "all of the v_n are integral" is false, but that the first counterexample is approximately 10^178485291567.
Solution

During the conference, a further problem was posed by one of the participants, namely, to prove that in any covering of a square checkerboard (of necessarily even edge length) by 2 x 1 dominos, there would always be a horizontal or vertical line through the checkerboard not passing through any domino.
During the ensuing discussion, a generalisation was conjectured by another participant (viz., me), namely, that the same is true for any rectangular checkerboard.
Since I know that I was not the only person who tried to prove these assertions, it may be worth communicating that both are in fact wrong, counterexamples of smallest size for the generalisation and for the original assertion being the 5x6 and 8x8 checkerboards:

    1 2 3 3 1 1              1 1 2 1 1 2 2 3
    1 2 1 2 2 3              2 3 2 3 3 4 1 3
    3 3 1 3 1 3     and      2 3 1 1 2 4 1 2  ,    respectively.
    4 2 2 3 1 2              1 4 4 3 2 3 3 2
    4 1 1 4 4 2              1 3 1 3 1 1 4 1
                             2 3 1 2 4 3 2 2
                             2 4 4 2 4 3 4 1
                             1 1 3 3 1 1 4 1