Simple unsolved math problem, 6

If you know a little point-set topology, below is an unsolved math problem whose statement is relatively simple.

Probably everyone has at least seen the Mandelbrot set in some form, as it’s a popular object of mathematical artists. Here’s a picture from Wikipedia:


The formal definition is as follows. Let f_c (z)=z^2+c, where c\in \mathbb{C} is a complex number. The Mandelbrot set X is the complex plot of the set of complex numbers c for which the sequence of iterates

f_c (0), f_c (f_c (0)), f_c (f_c (f_c (0))), \dots,

remains bounded in absolute value.
We say X is locally connected if every point x\in X admits a neighborhood basis consisting entirely of open, connected sets.

Conjecture: The Mandelbrot set X is locally connected.

Simple unsolved math problem, 5

It seems everyone’s heard of Pascal’s triangle. However, if you haven’t then it is an infinite triangle of integers with 1‘s down each side and the inside numbers determined by adding the two numbers above it:


First 6 rows of Pascal’s triangle

The first 6 rows are depicted above. It turns out, these entries are the binomial coefficients that appear when you expand (x+y)^n and group the terms into like powers x^{n-k}y^k:


First 6 rows of Pascal’s triangle, as binomial coefficients.

The history of Pascal’s triangle pre-dates Pascal, a French mathematician from the 1600s, and was known to scholars in ancient Persia, China, and India.

Starting in the mid-to-late 1970s, British mathematician David Singmaster was known for his research on the mathematics of the Rubik’s cube. However, in the early 1970’s, Singmaster made the following conjecture [1].

Conjecture: If N(a) denotes the number of times the number a > 1 appears in Pascal’s triangle then N(a) \leq 12 for all a>1.

In fact, there are no known numbers a>1 with N(a)>8 and the only number greater than one with N(a)=8 is a=3003.


[1] Singmaster, D. “Research Problems: How often does an integer occur as a binomial coefficient?”, American Mathematical Monthly, 78(1971) 385–386.

Simple unsolved math problem, 3

A perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself. For example,  1 + 2 + 3 = 6 implies 6 is a perfect number.

Unsolved Problem: Are there any odd perfect numbers? 

The belief, by some, that there are none goes back over 500 years (wikipedia).

If you want to check out some recent research into this problem, see


(Another unsolved problem: Are there an infinite number of even perfect numbers?)

Simple unsolved math problem, 2

In 1911, Otto Toeplitz asked the following question.

Inscribed Square Problem: Does every plane simple closed curve contain all four vertices of some square?

This question, also known as the square peg problem or the Toeplitz’ conjecture, is still unsolved in general. (It is known in lots of special cases.)


Inscribed square, by Claudio Rocchini

Thanks to Mark Meyerson (“Equilateral triangles and continuous curves”,Fundamenta Mathematicae, 1980) and others, the analog for triangles is true. For any triangle T and Jordan curve C, there is a triangle similar to T and inscribed in C. (In particular, the triangle can be equilateral.) The survey page by Mark J. Nielsen has more information on this problem.

Added 2016-11-23: See also this recent post by T. Tao.

Simple unsolved math problem, 1

In 1937 Lothar Collatz proposed the 3n+1 conjecture (known by a long list of aliases), is stated as follows.

First, we define the function f on the set of positive integers:

If the number n is even, divide it by two: f(n)=n/2.
If the number n is odd, triple it and add one: f(n)=3n+1.

In modular arithmetic notation, define the function f as follows:
f(n)=  {n/2},\  if \ n\equiv 0 \pmod 2, and f(n)=  {3n+1},\  if \ n\equiv 1 \pmod 2. Believe it or not, this is the restriction to the positive integers of the complex-valued map (2+7z-(2+5z)\cos(\pi z))/4.

The 3n+1 conjecture is: The sequence
n,\ f(n),\ f^2(n)=f(f(n)),\ f^3(n)=f(f^2(n)),\ \dots
will eventually reach the number 1, regardless of which positive integer n is chosen initially.

This is still unsolved, though a lot of people have worked on it. For a recent survey of results, see the paper by Chamberland.

Problem of the week, 161

A former colleague Bill Wardlaw (March 3, 1936-January 2, 2013) used to create a “Problem of the Week” for his US Naval Academy students, giving a prize of a cookie if they could solve it. One of them is given below.

The residue of an integer n modulo an integer d > 1 is the remainder r left when n is divided by d. That is, if n = dq + r for integers q and r with 0 < r < d, we write r \equiv n \pmod d for the residue of n modulo d. Show that the residue modulo 7 of a (large) integer n can be found by separating the integer into 3-digit blocks n = b(s)b(s-1)\dots b(1).(Note that b(s) may have 1, 2, or 3 digits, but every other block must have exactly three digits.) Then the residue modulo 7 of n is the same as the residue modulo 7 of b(1) - b(2) + b(3) - b(4) + \dots \pm b(s). For example,
n = 25,379,885,124,961,154,398,521,655 \pmod 7
\equiv 655 - 521 + 398 - 154 + 961 - 124 + 885 - 379 + 25 \pmod 7 \equiv 1746 \pmod 7 \equiv 746 - 1 \pmod 7 \equiv 745 \pmod 7 \equiv 3 \pmod 7.
Explain why this works and show that the same trick works for residues modulo 13.