Let s(n) be the sum of all positive divisors of n. E.g., s(2) = 1 + 2, s(4) = 1 + 2 + 4, s(20) = 1 + 2 + 4 + 5 + 10 + 20.
Let H(n) = 1 + 1/2 + 1/3 + 1/4 + ... + 1/n.
Question: is it true that s(n) < H(n) + log(H(n)) exp(H(n)) for all n > 1?
(log is natural logarithm)
It turns out that this simple looking problem is equivalent to the Riemann Hypothesis, which is perhaps the most important unsolved problem in pure mathematics. Many of the best minds in mathematics have attempted it in the approximately 160 years since Riemann posed it.
What equivalent means in this context is that if the Riemann Hypothesis is true, than the inequality above is also true, and if the inequality above is true, then the Riemann hypothesis is true.
For many problems one can easily guess the solution, without necessarily knowing the proof (like Fermat did). The answer is almost certainly negative, and a complete solution requires a very long and tedious proof that cannot be checked without a deep understanding of the field and many years of collective effort.
I am specifically interested in problems which can be understood by a child, have hard solutions which cannot be easily guessed, but are easily verified (ex. factoring the product of two large primes). It seems like these problems would make good candidates for one-way-functions, like the diophantine equations and ECC. But elliptic curve cryptology is too difficult to describe to a child.
You can even show how easy out would be for a simple curve where the left part is an easy equation.
The hard part is what makes an equation easy to solve. (And how some elliptic are broken.) It takes at least some high school math to hint at this.
What's mind blowing here is that if you ask for base 2, 3, 4 the answers are trivial but for bases 2,3,4,5 somehow you get 82000 and we do not know whether there is another and for 6 it's unknown.
Fermat's last theorem is an example, because it can be shown to be equivalent to, among other things, a very deep statement about elliptic curves (yep, the same thing The Fine Answer is about!) that Andrew Wiles proved.
I analyzed the problem for a while and theorized ways to collaboratively contribute to the solution. I settled on submitting an integer sequence to OEIS.
Solving them or progress? May be I don't care, but these little problem, simple ones does encourage one to start thinking about solving problems, they make math accessible to masses.
Here's one that's genuinely easy:
x^3 + y^3 + z^3 = 29
The solution is x = 1, y = 1, z = 3 (or any permutation thereof).So how about this one?
x^3 + y^3 + z^3 = 30
Play around a bit.Not so easy.
But it does have a solution!
Here it is:
x = -283059965
y = -2218888517
z = 2220422932
And what about x^3 + y^3 + z^3 = 33? There is no known solution. It's an open problem!This is a great visceral demonstration that Diophantene equations are hard: there can exist no algorithm that solves all of them. Diophantene equations, like the Halting Problem, are undecidable.
> 74 = −284650292555885^3 + 66229832190556^3 + 283450105697727^3 [1] [2].
Take any positive whole number, n.
If it is even, divide it by 2. (n = 2n)
If it odd, multiply it by 3 and add one. (n = 3n + 1)
Repeat. If n = 1, stop.
Does this process always lead to 1? It seems to, but who knows.
Does the fractional part of e^n tend to zero?