Can you give an example of a complex math problem that is easy to solve?

I am working on a project presentation and would like to illustrate that it is often difficult or impossible to estimate how long a task would take. I’d like to make the point by presenting three math problems (proofs, probably) that on the surface look equally challenging. But…

• One is simple to solve (or prove)
• One is complex to solve (or prove)
• And one is impossible

So if a mathematician can’t simply look at a problem and say, “I can solve that in a day, or a week, or a month, how can anyone else that is truly solving a problem? The very nature of problem solving is that we don’t know where the solutions lies and therefore we don’t know how long it will take to get there.

Any input or suggestions would be greatly appreciated.

This isn't exactly what you're asking for, but it should serve the same purpose very nicely.

Hilbert gave a talk in 1920 or so in which he discussed the difficulty of various problems.

He said that great progress had been made in analytic number theory in recent years, and he expected to live to see a proof of the Riemann Hypothesis.

Fermat's Last Theorem, he said, was harder; maybe the youngest members of his audience would live to see a proof.

But the problem of determining whether $2^{\sqrt2}$ is transcendental was so hard that not even the children of the youngest people in the audience would live to see a solution to that one.

With the benefit of hindsight, we can see that Hilbert had it exactly backwards.

$2^{\sqrt2}$ was settled in 1929 - Hilbert lived to see it.

Fermat, as we know, held out until the 1990s.

And the Riemann Hypothesis is still unsettled.

The point of the story is not to make fun of Hilbert. The point of the story is that if even Hilbert, the strongest mathematician of his era, could be so wrong in judging the relative difficulty of various mathematical problems, then it must be a really hard thing to do - which, I think, is the point you are trying to make.

For an integer $n$, let's seek integral solutions of $x^3 + y^3 + z^3 = n$.

1) When $n = 29$ a solution is easy to find: $(x,y,z) = (3,1,1)$.

2) When $n = 33$ it is harder to find a solution, but one is known: $$ (x,y,z) = (8866128975287528, -8778405442862239, -2736111468807040). $$ This was found in 2019 by Andrew Booker. See https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf and https://www.youtube.com/watch?v=ASoz_NuIvP0.

3) Here I had earlier used $n = 42$, saying we expect it is a sum of three cubes in $\mathbf Z$ but that no representation of $42$ in that form was currently known at the time I wrote that. Now (Sept. 2019) a representation is known: $42 = (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3$. I don't want to keep updating this answer again and again, so I'll just say we expect each integer $n \not\equiv 4, 5 \bmod 9$ is a sum of three cubes in $\mathbf Z$ and for such general $n$ this is still an open problem.

One set that (I think) meets your requirements and for which the statements are accessible to many:

1. Prove that $\sqrt{2}$ is irrational
2. Prove that $e$ is irrational
3. Prove that $e+\pi$ is irrational

A famous example, although it might require more background than you'd like, is Burnside's problem, which asks the following: if a group is finitely-generated and has the property that every element has order $n$ for some fixed positive integer $n$, is it necessarily finite?

• For $n = 2$ the answer is yes by a straightforward argument. If $a, b$ are two elements of the group, then $a^2 = b^2 = (ab)^2 = e$ implies $ab = ba$, so the group is abelian and has order dividing $2^m$ where $m$ is the number of generators.
• For $n = 3, 4, 6$ the answer is still yes, but the argument is more difficult.
• For $n = 5$ the problem is still open! (I think. According to Wikipedia, at the very least the specific case of two generators is still open.)