These are just nice things to think about. I don’t know why, but some of them are just nice.
I forget some of these, from time to time, so it’s nice to write them all down in one place. I may have made some mistakes paraphrasing them. In no particular order:
Goldbach’s conjecture
every whole number can be written as the sum of two half-primes (except 1)
Equivalently,
every whole number is the average of two primes (except 1)
every even whole number can be written as the sum of two primes (except 2)
Legendre’s conjecture
$N^{2} .. (N+1)^{2}$ contains a prime
Opperman’s conjecture
$N^{2} … N \cdot (N+1)$ contains a prime, as does $N \cdot (N+1) … (N+1)^{2}$
Andrica’s conjecture
$\sqrt{p^{N}} < 1 + \sqrt{p^{N-1}}$
so, for example, $\sqrt{7} < 1 + \sqrt{5}$, $\sqrt{11} < 1 + \sqrt{7} < 2 + \sqrt{5}$, &c &c
Firoozbahkt’s conjecture
$\sqrt[N+1]{p_{N+1}} < \sqrt[N]{p_{N}}$
so, for example, $\sqrt[4]{7} < \sqrt[3]{5} < \sqrt[2]{3}$
Pollock’s conjecture
Really going down a rabbit hole here
every whole number can be written as the sum of five tetrahedral numbers
Tetrahedral numbers are physically the number of oranges you can stack in a pyramid with a triangular base, N oranges wide
$$ Te_{1} = 1 \\ Te_{2} = 1 + (1 + 2) = 4 \\ Te_{3} = 4 + (1 + 2 + 3) = 10 $$The triangular layers form a sum of triangle numbers $T(N) = \frac{1}{2} N (N + 1)$:
$$ Te_{N} = \sum_{i=1}^{N} T(N) $$So if we have a bunch of triangles, and pair them with slightly smaller (or larger) triangles, the squares will complete, if you think about it:
$T_{N-1} + T_{N} = 1^{2} + … + N^{2}$
So they’re each roughly half a sum-of-squares in maybe the same sense that triangular numbers are each roughly half of one square.
Fermat’s polygonal number theorem
Proven, though I wouldn’t have expected it. Fermat, Lagrange, Gauss, and Cauchy all solved some piece of it.
every whole number can be written as a sum of n n-gonal numbers: 3 triangular numbers, 4 square numbers, 5 pentagonal numbers, ..
..which of course raises all sorts of eyebrows and squints towards Goldbach’s conjecture, as if half-primes were 2-sided/2-gonal numbers or something (?)
Twin prime conjecture
there are infinitely many pairs of primes N-1, N+1
Lehmer’s totient problem
So $\phi(n)$ is the totient of $n$, the number of coprimes below $n$. This is $p-1$ for a prime $p$, and is more like $\pi(n)$ for an $n$ with many factors.
Known:
$\phi(p)$ divides $p-1$
and conjectured:
$\phi(c)$ divides $c-1$ for some composite $c$
Carmichael’s totient problem
no whole number $n$ has a unique totient $\phi(n)$
In other words,
for every whole number $n$, there is another, $m$, such that $\phi(n) = \phi(m)$
Wolstenholme’s problem
Known, supposedly:
if $n$ is any prime number, $\dbinom{2n-1}{n-1} \equiv +1 \left( n^{3} \right)$
what’s conjectured is the other way around:
if $\dbinom{2n-1}{n-1} \equiv +1 \left( n^{3} \right)$, then $n$ must be prime
Collatz conjecture / Syracuse problem
In various flavors,
$$ c(n) = \begin{cases} 3n + 1 & \mathsf{n\ is\ odd} \\ \frac{1}{2}n & \mathsf{n\ is\ even} \end{cases} $$I like to just count upwards to a power of 2, though:
$$ 15 \cdot 3^{5} + 2^{0} \cdot 3^{4} + 2^{1} \cdot 3^{3} + 2^{2} \cdot 3^{2} + 2^{3} \cdot 3^{1} + 2^{8} \cdot 3^{0} = 4,096 $$where the idea is any odd number can be plugged in on the left as the leading coefficient of a $3^{x}$ polynomial, along with other coefficients of $2^{y}$ to all add up to some $2^{k}$.
How to look at an odd number and tell how it will decompose (or rather, how the power of 2 will decompose into the polynomial), without going ahead and doing the algorithm … well, that’s the problem.
Irrationals
A lot of these could go either way, I guess, but come on I think we all assume they’re at least irrational:
$$ \zeta(5) \ \text{is irrational, as well as any \& all} \ \zeta(5+2k) $$$$ \gamma \ \text{is irrational, algebraic, \&/ transcendental} $$abc conjecture
I vaguely kind of understand everything up to this point. I definitely don’t understand what’s below.
This one’s kind of a strange perspective: if we define some arbitrary “quality” to triplets of numbers, and only consider coprime triplets (no shared factors) as q(a, b, c)
$q(a, b, c) = \dfrac{\log{c}}{\log{ \mathsf{radical}(a \cdot b \cdot c) }}$
where, bear with me, the “radical” of abc is just the bottom row of primes, no exponents if you express a, b, c each as prime products; in better words,
$\mathsf{radical}(a \cdot b \cdot c) = \mathsf{the\ product\ of\ the\ distinct\ prime\ factors\ of\ n}$
So, we can finally say—err, conjecture—that, given the perspective that it’s rather hard to exceed quality 1.0, but it can be done
for every quality $1+ε$, there are only finitely many triples with $q(a, b, c) > 1+ε$
Let’s just focus around a rather fixed c value, and look at a, b pairs that could sum to roughly c. This means, for a fixed c, we aim to minimize $\mathsf{radical}(a \cdot b \cdot c)$.
…in other words, it’s rare to stumble upon two numbers a, b to be made of small prime powers that have a relatively large sum also made of small prime powers (different powers, all around), especially rare as we consider relatively larger and larger sums.
Even if you pick a, b to both be made of small primes, so as to minimize the rad(abc) on bottom (the c being mostly fixed), you still end up with enough big primes in c that it blows up the product. If you do find a (relatively) large c that can in fact be made of small primes, you won’t find many a, b pairs that are also made of small primes.
That’s the idea, I think. Just glancing around c ≅ 125:
128: $3 + 5^{3} = 2^{7}$, $\dfrac{log(128)}{log{(2 \cdot 3 \cdot 5)}} = (q = 1.42)$
131: $2 \cdot 3 + 5^{3} = 131$, $\dfrac{log(131)}{log{(2 \cdot 3 \cdot 5 \cdot 131)}} = (q = 0.59)$
132: $5^{3} + 7 = 2^{2} \cdot 3 \cdot 11$, $\dfrac{log(132)}{log{(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11)}} = (q = 0.63)$
133: $2^{3} + 5^{3} = 7 \cdot 19$, $\dfrac{log(133)}{log{(2 \cdot 5 \cdot 7 \cdot 19)}} = (q = 0.68)$
So it is rather unusual, like we should expect.
Birch-Swinnerton-Dyer conjecture
I’ll have to come back around to this one; I remember reading about it. It’s a bit hard to grasp, even the concept. There were a few minutes, maybe, where I understood it. That time has passed. Elliptic curves are kind of strange.
Riemann’s hypothesis
Ok, this is the one everybody knows and loves. We all want a proof. Let me just write out the badass mirror-image equation describing the complex extension, because who can remember that:
$$ \zeta(s) = \zeta(1 - s) · \Gamma(1 - s) · 2^{s} · \pi^{s-1} · sin(\frac{1}{2} \cdot s \cdot \pi) $$where $\Gamma(1 - s)$ is gamma, the complex extension of the factorial:
$$ \Gamma(x) = \int_{0}^{\infty} e^{-x} x^{z-1} \, dz $$The simplest way I know to think about it is with the Mertens function $M(n)$, which is “just” a running sum of the Mobius function1:
$$ M(n) = \sum_{k=1}^{n} \mu_{k} $$so the idea is we only care about the big-O of M(n):
$$ M(n) = O(n^{\frac{1}{2} + \epsilon})$$Supposedly this is equivalent. And the proof doesn’t work without the $\epsilon$.
Seems to me the bump in the rug is the Mobius function. There’s something going on with even-number-of-primes versus odd-number-of-primes, in a game where numbers-with-two-of-the-same-prime aren’t even playing. Strange. No idea if anyone is ever going to solve it.
Fermat’s last theorem
Famously proven by Andrew Wiles2,
$$ A^{N} + B^{N} \neq C^{N}, N \geq 3 $$Apparently, the connection to elliptic curves is that if we have some hypothetical solution $a$, $b$, $c$, & $n$ to the above, then we might want to care about the elliptic curve described by
$$ y^{2} = x \cdot (x - a^{n}) \cdot (x + b^{n}) $$or, supposedly equivalently, (?)
$$ y^{2} = x \cdot (x - a^{n}) \cdot (x - c^{n}) $$and as the story goes, this thing can’t be modular (whatever that is..3) if there really is some $a$, $b$, & $c$. But Wiles showed that every elliptic curve4 is modular, so that’s that.
Note: there have been so many errors in this post it’s crazy, I’ve rewritten it over the years (as blog rerolls go) from MathJax to HTML to Markdown to Markdown+HTML back to MathJax that at this point I should just rewrite it from scratch. I apologize for that. It’s a work in progress.