# Problem Sessions

One of the attractions of West Coast Number Theory are the problem sessions. The sessions are freeform. If you have a question, ask it. If you have an answer to a previous question, give it. Questions and answers are bundled up and distributed prior to the next conference.

A quick anecdote, via John Brillhart

At all the sessions of [West Coast Number Theory], Paul Erdős always sat in the middle of the front row. Often, no matter what was going on, he would seem to nod off and not pay attention. One never knew whether he were listening or not.

At the time, the problem sessions were run by John Selfridge, who would call on people to come up front to propose their problem . He always saved calling on Paul until the end when he had called on everyone who had raised their hand. He knew that Paul would have some problem to give.

One time at the point when no one had another problem to give, John turned to Paul who seemed to have dozed off and asked him: “Paul, do you have a problem?”

With this, Paul somewhat roused himself, and gesturing vaguely said, “I’ll be alright in a minute”.

Here are PDF’s of the problems sets from previous meetings.

### 7 responses to “Problem Sessions”

1. In 2012 West Coast Number Theory Conference problem sets, in problem 012.11, Mits Kobayashi [1] proposed the following unsolved problem:
For the divisor function σ, Is there any other solution to:
σ(pe) = σ(qf )
where p, q are distinct primes and e and f are integers greater than 1, other than σ(24) = σ(52)?
In this paper I give the answer ”no” assumming f = 2.

2. The case f = 2 is already covered by the Maohua Le paper mentioned in the remarks on 012:11.

3. I actually solved problem 0-16-13 and the answer is gcd(alpha, gamma) = [gcd(a, c)]^n.

Syrous Marivani

4. Sorry, the correct answer is gcd(alpha, gamma) =[gcd(a, bc)]^n.

• In 016:13, let a = b = c = 2, d = 1, n = 2. Then gcd(alpha, gamma) = gcd(8, 6) = 2, but (gcd(a, bc))^n = (gcd(2, 4))^2 = 2^2 = 4.

5. I also solved Problem 015-13. The answer is yes. There are infinitely many integers n such that n, n + 1, n + 2, n + 3, n + 4, and n + 5 are sums of two squares.

6. Syrous, at least one of those six numbers, n, n + 1, n + 2, n + 3, n + 4, n + 5, has to be 3 modulo 4 and therefore not a sum of two squares.