## Open problems with monetary rewards

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

Since the old days, many mathematicians have been used to attaching monetary rewards to problems they admit are difficult. Their reasons could be to draw other mathematicians' attention, to express their belief in the magnitude of the difficulty of the problem, to challenge others, "to elevate in the consciousness of the general public the fact that in mathematics, the frontier is still open and abounds in important unsolved problems.1", etc.

Current major instances are

Other problems with money rewards

-Kimberling's list of problems

Question: What others are there? To put some order into the answers, let's put a threshold prize money of 100 USD. I expect there are more mathematicians who have tucked problems in their web-pages with some prizes.

What this question does not intend to achieve:

-once offered but then collected or withdrawn offers

-new pledges of sums of money just here

P.S. Some may be intersted in the psychological aspects of money rewards. However, to keep the question focused, I hope this topic won't be ignited here. One more, I understand that mathematician's do not work merely for money.

-

3
I parse the title as "Difficulties arising by offering money for solutions to questions." Perhaps you might disambiguate the title? Gerhard "Offer Drinks Instead Of Money" Paseman, 2011.05.26 – Gerhard Paseman May 26 2011 at 18:38

6
I see some votes to close this post. I have started a meta asking why here: meta.mathoverflow.net/discussion/1055/… So, please all discussion concerning this should happen there. – To be cont'd May 27 2011 at 9:08

7
@Thierry -- I've often felt that even for the Clay million-dollar problems, that's got to be one of the hardest ways to earn a million dollars! I should think just the glory of solving such a problem would be, by far, a greater motivation. (BTW: I think it is small.) – Todd Trimble May 27 2011 at 9:40

12
Well, this question already has 4 votes to close. To whoever is thinking of pulling the plug: I also thought of voting to close at first, but since the question has already generated a couple of useful answers (including some problems I would probably never have heard of otherwise), I think it should stay open. – algori May 28 2011 at 1:57

11
I agree, I think that four votes to close is four too many. – Greg Kuperberg May 29 2011 at 13:42
show 7 more comments

## 19 Answers

Two which are for food rather than cash:

Let $f = t^{2d} + f_1 t^{2d-1} + f_2 t^{2d-2}+ \cdots f_d t^d + \cdots+ f_2 t^2 +f_1 t + 1$ be a palindromic polynomial, so the roots of $f$ are of the form $\lambda_1$, $\lambda_2$, ..., $\lambda_d$, $\lambda_1^{-1}$, $\lambda_2^{-1}$, ..., $\lambda_d^{-1}$. Set $r_k = \prod_{j=1}^d (\lambda_j^k-1)(\lambda_j^{-k} -1)$.

Conjecture: The coefficients of $f$ are uniquely determined by the values of $r_1$, $r_2$, ... $r_{d+1}$.

Motivation: When computing the zeta function of a genus $d$ curve over $\mathbb{F}_q$, the numerator is essentially of the form $f$. (More precisely, it is of the form $q^d f(t/\sqrt{q})$ for $f$ of this form.) Certain algorithms proceed by computing the $r_k$ and recovering the coefficients of $f$ from them. Note that you have to recover $d$ numbers, so you need at least $r_1$ through $r_d$; it is known that you need at least one more and the conjecture is that exactly one more is enough.

Reward: Sturmfels and Zworski will buy you dinner at Chez Pannise if you solve it.

Consider the following probabilistic model: We choose an infinite string, call it $\mathcal{A}$, of $A$'s, $C$'s, $G$'s and $T$'s. Each letter of the string is chosen independently at random, with probabilities $p_A$, $p_C$, $p_G$ and $p_T$.

Next, we copy the string $\mathcal{A}$ to form a new string $\mathcal{D}_1$. In the copying process, for each pair $(X, Y)$ of symbols in ${ A, C, G, T }$, there is some probability $p_1(X \to Y)$ that we will miscopy an $X$ as a $Y$. (The $16$ probabilities stay constant for the entire copying procedure.)

We repeat the procedure to form two more strings $\mathcal{D}_2$ and $\mathcal{D}_3$, using new probability matrices $p_2(X \to Y)$ and $p_3(X \to Y)$.

We then forget the ancestral string $\mathcal{A}$ and measure the $64$ frequencies with which the various possible joint distributions of `$\{ A, C, G, T \}$` occur in the descendant strings $(\mathcal{D}_1, \mathcal{D}_2, \mathcal{D}_3)$.

Our procedure depended on $4+3 \times 16$ inputs: the $(p_A, p_C, p_G, p_T)$ and the $p_i(X \to Y)$. When you remember that probabilities should add up to $1$, there are actually only $39$ independent parameters here, and we are getting $63$ measurements (one less than $64$ because probabilities add up to $1$). So the set of possible outputs is a semialgeraic set of codimension $24$.

Conjecture: Elizabeth Allman has a conjectured list of generators for the Zariski closure of the set of possible measurements.

Motivation: Obviously, this is a model of evolution, and one which (some) biologists actually use. Allman and Rhodes have shown that, if you know generators for the ideal for this particular case, then they can tell you generators for every possible evolutionary history. (More descendants, known sequence of branching, etc.) There are techniques in statistics where knowing this Zariski closure would be helpful progress.

Reward: Elizabeth Allman will personally catch, clean, smoke and ship an Alaskan Salmon to you if you find the generators. (Or serve it to you fresh, if you visit her in Alaska.)

-

2
Thanks for the nice exposition of the problems. – To be cont'd May 27 2011 at 8:45
Do you mean, independently at random rather than uniformly at random? – Greg Kuperberg May 27 2011 at 20:28
Corrected, thanks. – David Speyer May 27 2011 at 21:06
Is there some genericity condition on the $\lambda_i$ in the first conjecture? For instance if $\lambda_1 = 1$ then all the $r_k$ are zero, and that's obviously not enough information to reconstruct $f$? – Abhinav Kumar Jul 26 at 21:19
Yes, the conjecture is for generic $\lambda_i$. I can't seem to find an original print source for it, but you can find it restated as Conjecture 1.2 in arxiv.org/abs/math.AG/0411414 – David Speyer Jul 27 at 1:02

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

Addendum: There is a paper by Fan Chung similar to this book by Chung and Graham, Open problems of Paul Erdős in graph theory. She says there, "In November 1996, a committee of Erdős' friends decided no more such awards will be given in Erdős' name." But the same article says that Chung and Graham decided to still sponsor questions in graph theory, and this article in Science Magazine implies that they are still sponsoring the Erdős problems in general.

Some 19 years ago I collected a list of Erdős prize problems and posted them to Usenet. The problems were from "A Tribute to Paul Erdős" (1990) and "Paths, Flows, and VLSI Layout" (1980). I can repeat the problems here, although I have no idea which ones may have been solved.

$\$10000$. (T4N) Consecutive primes are often far apart. Conjecture: For every real number$C$, the difference between the$n\$'th prime and n+1'st prime exceeds

$$C \log(n) \log(\log(n))\log(\log(\log(\log(n))))/\log(\log(\log(n)))^2$$

infinitely often. (The wording in the source does not clearly indicate that the money will be awarded if the conjecture is disproved, only if it is proved.)

$\$3000\$. (T3N) Divergence implies arithmetic progressions. If the sum of the reciprocals of a set of positive integers is infinite, must the set contain arbitrarily long finite arithmetic progressions?

$\$1000$. (T2N) Unavoidable sets of congruences. A set of congruences$n = a_1 \bmod b_1$,$n = a_2 \bmod b_2$,... is unavoidable if each$n$satisfies at least one of them. Is there an$N$such that every unavoidable set of congruences either has two equal moduli$b_i$and$b_j$or some modulus$b_i$less than$N\$?

$\$1000$. (T1C) Three-petal sunflowers. Is there an integer$C$such that among$C^n$sets with$n$elements, there are always three whose mutual intersection is the same as each pairwise intersection? (Problem P2 is the same, except that Erdos asks about$k$-petal sunflowers for every$k$but then says he would be satisfied with$k=3\$.)

$\$500$. (T7N) Asymptotic bases of order 2 (I). Consider an infinite set of positive integers such that every sufficiently large integer is the sum of two members of the set. Can there be an$N$such that no positive integer is the sum of two members of the set in more than$N\$ ways?

$\$500$. (T8N) Asymptotic bases of order 2 (II). In the context of the previous problem, let$f(n)$be the number of ways that n is the sum of two members of the set. Can$f(n)/\log(n)$converge to a finite number as$n\$ goes to infinity?

$\$500$. (T9N) Evenly distributed two-colorings. Given a black-white coloring of the positive integers, let$A(n,k)$be the number of blacks minus the number of whites among the first$n$multiples of$k$. Can the range of$A\$ be bounded on both sides?

$\$500$. (T4C) Friendly collections of half-sized subsets. Given$1+(\binom{4n}{2n} - \binom{2n}{n}^2)/2$distinct, half-sized subsets of a set with$4n\$ elements, must there be two subsets which intersect only in one element? (As problem P1, 250 pounds is offered.)

$\$500$. (T1G) Uniformity of distance in the plane (I). Is there a real number$c$such that n points in the plane always determine at least$cn/\sqrt{\log(n)}\$ distinct distances?

$\$500$. (T1G) Uniformity of distance in the plane (II). Is there a real number$c$such that given n points in the plane, no more than$n^{(1+c/\log(\log(n)))}\$ pairs can be unit distance apart?

$\$500$. (P2) Sets with distinct subset sums. Is there a real number$c$such that, given a set of n positive integers whose subsets all have distinct sums, the largest element is at least$c2^n\$? (As in problem T1N, no prize is mentioned.)

$\$250$. (P4) Collections of sets not represented by smaller sets. Is there a real number$c$such that for infinitely many positive integers$n$, there exists$cn$or fewer sets with n elements, no two of which are disjoint, and every$(n-1)\$-element set is disjoint from at least one of them?

$\$250/\$100$. (P15) Slowly increasing Turan numbers. If H is a (simple) graph, the Turan number $T(n,H)$ is the largest number of edges a graph with $n$ vertices can have without containing a copy of $H$. Conjecture: the function $f(n) = T(n,H)/n^{3/2}$ is bounded above if and only if every connected subgraph of $H$ has a vertex of valence 1 or 2. The larger award would be granted for a proof.

$\$100/\$25000$. (T6N) Consecutive early primes. An early prime is one which is less than the arithmetic mean of the prime before and the prime after. Conjecture: There are infinitely many consecutive pairs of early primes. The larger award would be granted for a disproof.

$\$100$. (T8G) Quadrisecants in the plane. Given an infinite sequence of points in the plane, no five of which are collinear, let$r(n)$be the number of lines that pass through four points among the first$n$. Can it happen that$r(n)/n^2\$ does not converge to zero?

-

@Greg, thank you very much. Your collection of the Erdos prize problems were linked to previously but this organized answer makes the question really interesting.  – To be cont'd May 28 2011 at 5:47

Of course mathematicians do not work merely for money! They are motivated by higher things like access to moderation tools at 10000 reputation level on Mathoverflow!

I've heard from Ron Graham that the bookkeeping for Erdos rewards is difficult because many people frame the check and never cash it.

There is always the chance of earning \$327.68 from Donald Knuth. It is stretching things more than a bit to include that in and of itself, but the linked article is amusing and the general considerations are pertinent.

The EFF offers large rewards for large primes.

-

10
I've heard that the cleverer people who win Erdos money photocopy the check, frame the copy, and keep the original. – Gerry Myerson May 27 2011 at 5:37

6
Knuth says his cheques are much more often cached than cashed. – Max Jun 21 2011 at 19:09

Paul Erdős was famous for offering money for solutions to math problems. My understanding is that those prizes are still being administered by Ron Graham, even though Erdős passed away several years ago.

One of his most famous (and biggest money) problems is the following:

Conjecture: If $S$ is a set of positive integers such that the series $\sum_{s \in S} \frac{1}{s}$ diverges, then $S$ contains arbitrarily large arithmetic progressions.

I believe that \$3000 is offered for a proof.

Several graph theory related problems, with prize money listed for many of them, are collected in Erdős on graphs, by Chung and Graham.

-
@Matthew, thank you. – To be cont'd May 26 2011 at 18:20

7
With a solution of that problem a Field medal would also come, unless you are over 40. – Maurizio Monge May 26 2011 at 18:48

1
See "Does there exist a comprehensive compilation of Erdos open problems?" mathoverflow.net/questions/27716 – Joseph O'Rourke May 26 2011 at 20:52

3
A partial list of problems with monetary awards is available at: math.niu.edu/~rusin/known-math/93_back/prizes.erd – Ben Linowitz May 27 2011 at 1:16

9
@Maurizio: I'm wondering what makes you so sure of that? – Todd Trimble May 27 2011 at 21:47
show 2 more comments

See http://people.math.jussieu.fr/~talagran/prizes.pdf for a total of $\$7000$offered by Michel Talagrand for a solution of three problems. In particular$\$5000$ for the "Bernoulli conjecture". You may also be interested in the following picture of Mazur awarding Per Enflo a live goose as promised for the solution of the approximation problem. http://en.wikipedia.org/wiki/File:MazurGes.jpg

-

John Conway has prize money out for various problems, such as \$1000 for the thrackle conjecture.

-

It was pointed out by Randall Munroe that by proving the inconsistency of logic you can earn quite a lot of money:

(Source)

-
Ex falso quodlibet! So, I doubt this creates that many errors. – quid Jul 26 at 12:00

9
I should have mentioned that the hidden text is : "Dear Reader: Enclosed is a check for ninety-eight cents. Using your work, I have proven that this equals the amount you requested." – Adrien Jul 26 at 12:21
Oh, so you suggest to prove just this, and then cash in essentially all the things mentioned so far. Nice plan ;D – quid Jul 26 at 15:07

J.-B. Zuber offered respectively 1, 2 and 3 bottles of Champagne for the classification of finite quantum subgroups of $SU_q(3)$, $SU_q(4)$, and $SU_q(5)$, respectively.

Ocneanu solved the first two cases, but as far as I know the problem for $SU_q(5)$ is still open.

-

I am not sure if he would still be willing to pay out, but Vaughan Pratt offers 2000 smackers for the solution to the following:

http://thue.stanford.edu/puzzle.html

Let D be a set of subsets of a set A with the following three properties.

(i) D contains A and the empty set.

(ii) Let C be any set of pairs (a,b), a,b in A, such that for all a in A, the sets {b | (a,b) is in C} and {b | (b,a) is in C} are in D. Then {b | (b,b) is in C} is in D.

(iii) (T1) For any two distinct elements a,b of A, D contains a set containing a but not b, and another containing b but not a.

The set D consisting of all subsets of A evidently satisfies (i)-(iii). Does any other set of subsets of A do so?

-

1
If smackers is a synonym for kisses, I'm not so sure I want to work in the same field as this problem. Gerhard "Choosy About Who He Kisses" Paseman, 2011.05.26 – Gerhard Paseman May 27 2011 at 0:05
According to thefreedictionary.com, smacker = \$1 or £1. Which is a relief, 'cos my first thought was that it had something to do with smack. Which would be scandalous, of course. ;) – Nikita Sidorov May 27 2011 at 1:25
The free dictionary got it right. – Steven Gubkin May 27 2011 at 1:40
HAHAHA! This made me laugh. – Koundinya Vajjha May 27 2011 at 5:16

A. Bressan has advertised two monetary rewards of \$500 each for solutions to problems on mixing flows and blocking problems.

The first problem was unsolved as of Jan. 15, 2011, although progress in relevant directions is noted in the linked announcement. The second problem was announced on Jan. 19, 2011.

-

The $3n+1$ Conjecture has some money assigned to it.

Define $T(n) = n/2$ when $n$ is even and $3n+1$ when $n$ is odd.

For any positive integer $n$ does there exist a positive integer $N$, such that $T^N(n) = 1$?

The origin of this precise question seems to be obscure, although Lothar Collatz made similar conjectures during the 1930s1. For example, Bryan Thwaites claims to have been the first to make this conjecture in 19522, and this does not seem to have been decisively refuted. (The 1937 dates in the Wikipedia and Mathworld articles are missing citations - the Wikipedia edit dates to 7 September 2004.)

Rewards offered to date include 1000 UK pounds from Bryan Thwaites, 500 US dollars from Paul Erdos, and 50 dollars (Canadian?) from H.S.M. Coxeter1.

1. Lagarias, The $3x+1$ problem and its generalizations Am. Math. Monthly 92 (1985) 3-23.
2. Bryan Thwaites, Two conjectures or how to win £1100. Math. Gazette 80 (1996) 35-36.
-

3
The statement is not correct as $N$ must be allowed to depend on $n$. Is there documentation for Halmos' offer? Did the offer expire with the professor? Not that I expect anyone to collect on it any time soon.... – Gerry Myerson May 27 2011 at 5:41
As the conjecture is stated, False as $\forall N$, $T^N(2^{N+1}) = 2 \neq 1$ but I doubt I'll get \\$500 for stating that. – qwerty1793 May 27 2011 at 9:45

2
500 double dollars. I feel like I just walked into a trigun episode. – Steven Gubkin May 27 2011 at 14:09
@Gerry Thanks for pointing out my errors. – Daniel Parry May 27 2011 at 18:42

1
@Daniel, ordinarily I'd say, if Lagarias says so, it must be true, but I'm disappointed that he doesn't give any reference. – Gerry Myerson May 28 2011 at 5:49
show 2 more comments

Gerard Cornuejols offers \$5000 for the first proof (or refutation) of each of the 18 conjectures in his 2001 book "Combinatorial Optimization: Packing and Covering". Six of the conjectures have been resolved so far, five - by Maria Chudnovsky, Paul Seymour and coauthors.

-

Guy, Unsolved Problems In Number Theory, 3rd edition, A12 (page 45) says "Selfridge, Wagstaff & Pomerance offer 500.00 + 100.00 + 20.00 for a composite $n\equiv3{\rm\ or\ }7\pmod{10}$ which divides both $2^n-2$ and the Fibonacci number $u_{n+1}$ or 20.00 + 100.00 + 500.00 for a proof that there is no such $n$." John Selfridge having left us, I do not know whether his part of the offer still applies.

-

1
"John Selfridge having left us..." Oh dear, I didn't know that. He was my office-mate for a while when I was at Macquarie, and we had a lot of fun together. RIP... – Todd Trimble May 28 2011 at 11:11

I remember reading in Havil's book Gamma that supposidly Hardy was willing to offer his Savilian Chair at Oxford University to anyone who could prove that the Euler Mascheroni constant is irrational. I wonder if this offer still stands?

-

2
Page 52 of Havil's book, but no reference given there. Did Hardy really make this offer, even in jest? – Gerry Myerson May 30 2011 at 0:28

Bill Gasarch is offering \$289 for a 4-coloring of a 17-by-17 grid such that there is no rectangle with all four corners the same color. (No reward for proving one doesn't exist.)

Edit: The above problem, of 4-coloring grids, has since been solved completely; see the comments for the 12x21 case.

Scott Aaaronson offers $200 for an oracle relative to which BQP is not contained in PH, or$100 for an oracle relative to which BQP is not contained in AM.

-

Recently Ian Morrison issued a 100 dollar prize for the construction of an effective divisor on $\overline M_g$ with slope less than 6 (See the recent preprint of Chen, Farkas and Morrison).

-

As it is pointed out in a footnote in A. Zorich's survey "Flat surfaces", Anatole Katok promised (on behalf of the Center for Dynamics and Geometry of Penn State University) a prize of 10,000 euros for the solution of the problem of finding periodic orbits and describing the behavior of generic orbits of billiards in (all/almost all) triangular tables. See page 13 of the ArXiv version of the survey for more comments.

-

Stepan Holub is offering €200 for a solution to the following problem:

Do there exist $n\ge 2$ and nonempty words $u_1,\ldots,u_n$ such that

$(u_1 \ldots u_n)^2 = u_1^2 \ldots u_n^2$ and

$(u_1 \ldots u_n)^3 = u_1^3 \ldots u_n^3$

but the $u_i$ don't all commute with one another?

(I guess the $n\ge 2$ and nonemptiness requirement are technically redundant.)

Source: https://www.student.cs.uwaterloo.ca/~cs462/openproblems.html

-

I'm a computer scientist by trade, but I really just enjoy working on open mathematical problems in my free time. Kimberling's page is pretty nice as you mentioned, I was able to knock one of them out (a minor one, the Swappage Problem), and always look forward to when new ones are posted.

One of the best places I have found for open problems is probably open problem garden. The site is frequently updated with new problems ranging from graph theory, theoretical computer science, algebra, etc. The nice thing is that the problems are also ranked by relative difficulty.

As for cash: A number of the problems DO offer some type of cash bounty (as clearly indicated in the summary section for the problem next to "Prize" text if it exists). Problems such as The Erdos-Turan conjecture on additive bases offer cash incentives for solving.

There are also other problems listed that offer monetary compensation and are posted periodically throughout the site. However, sifting through the problems can be time consuming as many of them do not offer cash incentives.

-

4
If you can edit the answer so that it properly answers the question, please do so. Other MathOverflow questions talk about open problem lists; this one talks about those with a cash bounty (historical or present). Gerhard "Email Me About System Design" Paseman, 2011.06.20 – Gerhard Paseman Jun 21 2011 at 4:29
While not all the problems listed here offer cash bounty, a number of them do. I'll rephrase the post accordingly. – Vincent Russo Jun 21 2011 at 16:54
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 158, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9418177008628845, "perplexity_flag": "middle"}