> Mathematicians know that you can make problems arbitrarily complex, and declaring problems with large prizes attached to them can lead to a lot of competition

> Mathematicians know that you can make problems arbitrarily complex, and declaring problems with large prizes attached to them can lead to a lot of competition and drama. Yes, you can make problems arbitrarily complex. But the prize problems were chosen not just because the solutions appear likely to be very complex (the problem statements aren't necessarily inherently complex--there is a way to restate the Riemann hypothesis that a junior high school student could easily understand, which I'll give below). They were chosen because they were important problems that mathematicians really wanted solved, top people had worked on them for a long time and progress stalled a long time ago, and it seemed likely that solving them would require major breakthroughs. Those kind of problems can be discouraging. Enough people who are probably better than you have spent enough time failing to solve them that realistically most researchers are going to focus all their efforts on something they are likely to make progress on. A nice prize can get more people to at least work on them as side projects. Here's that restatement of the Riemann hypothesis I mentioned. The Riemann hypothesis is that the non-trivial zeros of the function ζ(s) occur on the line 1/2 + yi. ζ(s) is 1/1^s + 1/2^2 + 1/3^s + ... when s is a complex number whose real part is greater than 1, and defined everywhere else except s = 1 by a process called analytic continuation. The trivial zeros are at s = -2, -4, -6, ... . For a mathematician, or a non-mathematician who has taken complex analysis and hasn't forgotten much of that, that is not too complex a definition. For anyone else the first reaction is probably "Trivial zeros? How the heck does that thing even have zeros? And if it does how the heck can it have zeros at any negative integers! It is obviously infinity at every negative integer!!!". Here's a different hypothesis that turns out to be exactly equivalent to the Riemann hypothesis. They are either both true of both false, so resolving one of them resolves the other. Let H(n) = 1 + 1/2 + ... + 1/n for all positive integers n. These are called the harmonic numbers. Let S(n) = the sum of the positive integer factors of n for all positive integers n. For example S(4) = 1 + 2 + 4, S(6) = 1 + 2 + 3 + 6, and S(17) = 1 + 17. Hypothesis: S(n) <= H(n) + exp(H(n)) log(H(n)) with equality only when n = 1. The proof that this is equivalent to the Riemann hypothesis is here [1]. [1] https://arxiv.org/pdf/math/0008177 https://arxiv.org/pdf/math/0008177