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:
every whole number can be written as the sum of two half-primes (except 1)
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)
N2 .. (N+1)2 contains a prime
N2 .. N(N+1) contains a prime, and N(N+1) .. (N+1)2 contains another
√pN < 1 + √pN-1
so, for example, √7 < 1 + √5, √11 < 1 + √7 < 2 + √5, &c &c
N+1√pN+1 < N√pN
so, for example, 3√7 < 2√5
Really going down a rabbit hole here
every whole number can be written as the sum of five tetrahedral numbers
Remember, tetrahedral numbers are physically the number of oranges you can stack in a pyramid with a triangular base, N oranges wide
TN-1 + TN = 12 + .. + N2
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 (?)
Hurwitz quaternion norms
This isn’t really a theorem but common knowledge / simple algebra: if a Hurwitz quaternion q is either a 4-tuple of integers or a 4-tuple of half-integers (n+½),
the norm of any Hurwitz quaternion q = [a b c d] or q = [a½ b½ c½ d½] is a whole number
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) = log( c ) / log( rad(abc) )
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,
rad(n) = 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 rad(abc).
…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:
3 + 53 = 27 (128), rad(abc) = 2·3·5 « c (q = 1.42)
2·3 + 53 = 131, rad(abc) = 2·3·5·131 » c (q = .59)
7 + 53 = 22·3·11 (132), rad(abc) = 3·5·7·11 » c (q = .69)
23 + 53 = 7·19 (133), rad(abc) = 2·5·7·19 » c (q = .68)
So it is rather hard, like we should expect.
Twin prime conjecture
there are infinitely many pairs of primes N-1, N+1
Lehmer’s totient problem
So φ(n) is the totient of n, the number of coprimes below n. This is p-1 for a prime p, and is more like π(n) for a many-factored n.
φ(n) divides n-1 for all prime n
φ(n) divides n-1 for some composite n
Carmichael’s totient problem
no whole number n has a unique totient φ(n)
In other words,
for every whole number n, there is at least one other whole number m such that φ(n) = φ(m)
if n is any prime number, (2n-1 choose n-1) ≡ +1 (n3)
what’s conjectured is the other way around:
if (2n-1 choose n-1) ≡ +1 (n3), then n must be prime
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. Elliptic curves are kind of strange things.
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:
ζ(s) = ζ(1 - s) · Γ(1 - s) · 2s · πs-1 · sin(½·s·π)
where Γ(1 - s) is gamma, the complex extension of the factorial:
Γ(x) = ∫ xz-1 e-x dz over 0..∞
The simplest way I know to think about it is with the Merten’s function, which is just a running sum of the Mobius function1
M(n) = ∑ 1..n µ(k)
so the idea is we only care about the big-O of M(n):
M(n) = O(n ½ + ε )
Supposedly this is equivalent. And the proof doesn’t work without the ε.
Obviously, 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. Proving this kind of thing feels like moving a couch while you’re sitting on it2.
Collatz conjecture / Syracuse problem
In various flavors,
c(n) = 3n + 1 if n is odd, n/2 if n is even. c ∞(n) = 1
You can also just look at odd numbers, that’s for some reason called the Syracuse function, a function from odd numbers to other odd numbers, f(n) = (3n + 1) ⁄ 2a, for maximum a:
f ∞(n) = 1
I like to just count upwards to a power of 4(?), though:
15, 46, 140, 424, 1280, 4096
So, the way I like to see it, if you triple 15 enough times, and you add in the right powers of 2 (in increasing size) along the way (each power of 2 tripling a certain amount, always less than the original 15), then you can end up at a power of 4. Of course there’s no choice in Collatz but I like to view it as a game:
15·3 + 1 = 46
46·3 + 2 = 140
140·3 + 4 = 424
424·3 + 8 = 1,280
1,280·3 + 256 = 4,096
or in other words,
15·35 + 1·34 + 2·33 + 4·32 + 8·31 + 256 = 4,096
15·35 + 20·34 + 21·33 + 22·32 + 23·31 + 28 = 4,096
where the idea is any odd number can be plugged in on the left, and some sequence of powers of 2, enough of them (all increasing, skips are allowed) will get you to a power of 2. Any given power of 2 only has so many numbers that point to it, but an arbitrary large odd number on the left presumably will point to some 2k.
N·3a + 2k1·3a-1 + 2k2·3a-2 + … + 2ka-1·30 + 2ka = 2x
ß·3a + 2A·3a-1 + 2B·3a-2 + … + 2Y·31 + 2Z = 2Ω
A < B < .. < Z < Ω
so if you just write the above in base three, it’s
2Ω . = ß 2A 2B … 2Y 2Z .
How to look at an odd number and tell how it will decompose, without going ahead and doing the algorithm, is the heart of the problem, to me.
A lot of these could go either way, I guess, but come on I think we all assume they’re irrational:
ζ(5) is irrational
ɣ is irrational
ɣ is algebraic
ɣ is transcendental
Fermat’s last theorem
Famously proven by Andrew Wiles3,
AN + BN ≠ CN, N ≥ 3
Apparently, the connection to elliptic curves is that if we have some hypothetical solution a, b, c & n to the above,
an + bn = cn
then we might want to care about the elliptic curve described by
y2 = x·(x - an)·(x + bn)
or, supposedly equivalently,
y2 = x·(x - an)·(x - cn)
Let me just rant for a minute: theorems that depend on base-10 are generally not my thing. Adding digits together is like Boiling A Steak; I kinda “get it”, but I think it just kinda sucks. Different-base representations are interesting, but if you stick only only only to base-10 I don’t even bother6.
which you just gotta know, for now, I can’t explain it fast ↩
especially in the sense where I hope somebody does it for me, gently.. ↩
and not-quite-understood by almost every other human being on the planet, yours truly included ↩
Ⓧ Enter Nightmare Fog ↩
talk about overkill, like Yamato-cannons-on-zerglings-style show of force ↩
especially after having seen dozenal, and base-6 ↩