The Question :
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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.
The Question Comments :
The Answer 1
266 people think this answer is useful
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.
The Answer 2
91 people think this answer is useful
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.
The Answer 3
52 people think this answer is useful
One set that (I think) meets your requirements and for which the statements are accessible to many:
 Prove that $\sqrt{2}$ is irrational
 Prove that $e$ is irrational
 Prove that $e+\pi$ is irrational
The Answer 4
42 people think this answer is useful
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 finitelygenerated 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.)
The Answer 5
25 people think this answer is useful
It’s easy to multiply two 20digit prime numbers, but what if you had to factor their product? There you have an easy problem and a hard problem.
The Answer 6
21 people think this answer is useful
Take a look at the suggestions and the comments of each one.
Suggestions:
1 – $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$, Solve for $x$.
2 – Poincaré conjecture (Every simply connected, closed 3manifold is homeomorphic to the 3sphere.).
3 – The Continuum Hypothesis (First Hilbert problem). This question is indecidible.
Comments:
1 – This problem looks a very hard one. How to find $x$ when we raised it to it infinite times and this should be equal to $2$.
The truth is this problem could be solved on a “easy” way. See this my own question.
2 – This question is a very hard one. It has proved by Grigori Perelman. See this link for more details.
3 – This question is “impossible” because is indecidible and find a final answer isn’t possible. Reference.
The Answer 7
12 people think this answer is useful
Irrationality of the Riemann zeta function at integer values.
$\zeta(2)$ is irrational. This won’t be an obvious fact to those without a mathematics background, but most mathematicians are familiar with a proof that $\zeta(2)=\frac{\pi^2}{6}$. (And there are similar statements for the zeta function of even integers).
$\zeta(3)$ is irrational. Known, but not until 1978. Here we have irrationality, but no closed expression in terms of wellunderstood constants. (Nor, to my knowledge, is there any reason to expect such an expression to exist).
$\zeta(5)$ is irrational. Open. We know that $\zeta(2k+1)$ is irrational infinitely often, and least one of $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, and $\zeta(11)$ is irrational.
(See this post for links to more details. The Wikipedia article on Apéry’s Theorem also lists some results.)
The Answer 8
8 people think this answer is useful
What positive integers can be written as the sum of two squares? Sum of three squares? Sum of four?
For two squares, it’s all positive integers of the form $a^2b$, where $b$ isn’t divisible by any prime of the form $4k+3$, and the proof is easy.
For four squares, it’s all positive integers, and the proof is moderately difficult, but covered in any course on number theory.
For three squares, it’s positive integers not of the form $4^a (8b+7)$ and, while not impossible, the proof is considerably more difficult than the previous two.
The Answer 9
7 people think this answer is useful
Sometimes a problem which seems very hard turns out to be “easy” because someone was clever enough to look at it in just the correct way. Two examples of this:
a. Construct lots of nonhomiltonian planar 3valent 3connected graphs (such graphs can be realized by convex 3dimensional polyhedra) The Grinberg condition provides a nifty approach to finding such graphs easily: http://en.wikipedia.org/wiki/Grinberg%27s_theorem
b. Klee’s art gallery problem: find the number of vertex guards that are sometimes necessary and always sufficient to “see” all of the interior of a plane simple polygon with n vertices. V. Chvatal and Steve Fisk found simple ways to answer this question: http://en.wikipedia.org/wiki/Art_gallery_theorem
The Answer 10
4 people think this answer is useful
Have you considered the famous classification of all finite simple groups? See here for the history of this effort. In addition and intimately related, the conjecture of Burnside (1911): every finite group of odd order is solvable (or equivalently: every finite simple group has even order), which was settled by Walter Feit and John Thompson in 1963.
The Answer 11
4 people think this answer is useful
Let $D(M)=$ the collection of Dedekind cuts in the structure $M=(X,<)$.
(a) $D(\mathbb{Q})=\mathbb{R}$ (this is relatively easy)
(b) For any dense linear order $M=(X,<)$, $2^{M} \geq D(M)$ (this one is a little more challenging)
(c) For any dense linear order $M=(X,<)$, $2^{M}=D(M)$ (this one is impossible, since it is independent of ZFC).
The Answer 12
3 people think this answer is useful
This famous pell equation $$\Large{ 61 x^{2} + 1 =y^{2}}$$ has an trivial solution $(0,\pm{1})$. But if you were to look out for positive integral solutions then the smallest positive integral solution is given by $(226153980,1766319049)$. More on this here at this link
The Answer 13
3 people think this answer is useful
1) There is an infinite number of integer solutions to $a^2+b^2=c^2$ with $abc \neq 0$
2) There are no integer solutions to $a^n+b^n=c^n$ with $abc \neq 0$ and $n \geq 3$
3) If $a^x+b^y=c^z$ where $a,b,c,x,y,z$ are positive integers with $x,y,z >2$ then $a$, $b$ and $c$ have a common prime factor.
The Answer 14
1 people think this answer is useful

$\zeta(2) = \sum_n \frac{1}{n^2} = \frac{\pi^2}{6}$

$\zeta(1+it) \neq 0$

$\zeta(s)=0$ $\implies$ $\Re(s)= 1/2$, if $s$ is lies in the right halfplane.