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Problems : AMS Tex file or dvi file or postscript file

Solutions: AMS Tex file or dvi file or postscript file

1. Somebody incorrectly remembered Fermat's little theorem as saying that the congruence a^{n+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 2^{n} k+1.

Solution

3. Show that if (a^{2} + b^{2})/(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 (a_{1}, a_{2}, a_{3}, ... ) 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 x^{3} + 4 x^{2} + 3 for the prime 9403). Show that this polynomial is always irreducible.

Solution

2. Suppose that b_{n} x^{n} + b_{n-1} x^{n-1} + ... + b_{1} x + b_{0} in **R**[x] has only real roots. Show that the polynomial b_{n}/n! x^{n} + b_{n-1}/(n-1)! x^{n-1} + ... + b_{1}/1! x + b_{0}, 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 t_{1}, t_{2}, t_{3}, ... 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 u_{1}, u_{2}, u_{3}, ... 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 v_{1}, v_{2}, v_{3}, ... by the recursion

v_{n} =(2+v_{1}^{2} + ... + v_{n-1}^{2})/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

If you have any comments, you can mail me.

Don Zagier 1996