Some Favorite Math Theorems
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 halfprimes (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 .. (N+1)^{2} contains a prime
Opperman’s conjecture
N .. N(N+1) contains a prime, and N(N+1) .. (N+1)^{2} contains another
Andrica’s conjecture
√p_{N} < 1 + √p_{N1}
Firoozbahkt’s conjecture
^{N+1}√p_{N+1} < ^{N}√p_{N}
Pollock’s conjecture
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
T_{N1} + T_{N} = 1^{2} + .. + N^{2}
So they’re each roughly half a sumofsquares 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 ngonal 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 halfprimes were 2sided/2gonal 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 4tuple of integers or a 4tuple of halfintegers,
the norm of any Hurwitz quaternion q = [a b c d] or q = [a.5 b.5 c.5 d.5] is a whole number
abc conjecture
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 + 5^{3} = 2^{7}, rad(abc) = 2·3·5 « c (q = ~1.42)
2·3 + 5^{3} = 131, rad(abc) = 2·3·5·131 » c (q = ~.59)
7 + 5^{3} = 2^{2}·3·11, rad(abc) = 3·5·7·11 » c (q = ~.69)
2^{3} + 5^{3} = 7·19, 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 N1, N+1
Lehmer’s totient problem
So φ(n) is the totient of n, the number of coprimes below n. This is p1 for a prime p, and is more like π(n) for a manyfactored n.
Known:
φ(n) divides n1 for all prime n
and conjectured:
φ(n) divides n1 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)
Wolstenholme’s problem
Known, supposedly:
if n is any prime number, (2n1 choose n1) ≡ +1 (n^{3})
what’s conjectured is the other way around:
if (2n1 choose n1) ≡ +1 (n^{3}), then n must be prime
BirchSwinnertonDyer 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. Elliptic curves are kind of strange things.
Riemann’s hypothesis
Ok, this is the one everybody knows and loves. We all want a proof. Let me just write out the badass mirrorimage equation describing the complex extension, because who can remember that:
ζ(s) = ζ(1  s) · Γ(1  s) · 2^{s} · π^{s1} · sin(½·s·π)
where Γ(1  s) is gamma, the complex extension of the factorial:
Γ(x) = ∫ x^{z1} 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 function^{1}
M(n) = ∑_{ 1..n } µ(k)
so the idea is we only care about the bigO 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 evennumberofprimes versus oddnumberofprimes, in a game where numberswithtwoofthesameprime aren’t even playing. Proving this kind of thing feels like moving a couch while you’re sitting on it^{2}.
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) ⁄ 2^{a}, 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,
15·3^{5} + 1·3^{4} + 2·3^{3} + 4·3^{2} + 8·3^{1} + 256 = 4,096
or
15·3^{5} + 2^{0}·3^{4} + 2^{1}·3^{3} + 2^{2}·3^{2} + 2^{3}·3^{1} + 2^{8} = 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 4. Any given power of 4 only has so many numbers that point to it, but an arbitrary large odd number on the left presumably will point to some 4^{k}.
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.
Mahler’s 3⁄2 conjecture
where { .. } is the fractional part of the number:
{x·(3⁄2)^{n}} for all n, for some real x
The idea is there aren’t any of these numbers (no such x), but it’s hard to prove that.
Irrationals
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 Wiles^{3},
A^{N} + B^{N} ≠ C^{N}, N ≥ 3
Apparently, the connection to elliptic curves is that if we have some hypothetical solution a, b, c & n to the above,
a^{n} + b^{n} = c^{n}
then we might want to care about the elliptic curve described by
y^{2} = x·(x  a^{n})·(x + b^{n})
or, supposedly equivalently,
y^{2} = x·(x  a^{n})·(x  c^{n})
and as the story goes, this thing can’t be modular (whatever that is..^{4}) if there really is some a, b, c. But Wiles showed that every elliptic curve^{5} is modular, so that’s that.
Pet peeves
Let me just rant for a minute: theorems that depend on base10 are generally not my thing. Adding digits together is like Boiling A Steak; I kinda “get it”, but I think it just kinda sucks. Differentbase representations are interesting, but if you stick only only only to base10 I don’t even bother^{6}.

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 notquiteunderstood by almost every other human being on the planet, yours truly included ↩

Ⓧ Enter Nightmare Fog ↩

talk about overkill, like Yamatocannonsonzerglingsstyle show of force ↩

especially after having seen dozenal, and base6 ↩