this post was submitted on 15 Oct 2023
15 points (94.1% liked)

Monero Memes

292 readers
8 users here now

founded 1 year ago
MODERATORS
 

Hamilton was an Irish mathematician, who discovered quaternions on the 16th of October, 1843. When he discovered them, he was so happy that he carved his fundamental equations i² = j² = k² = ijk = −1 into the stone of a bridge (apparently he was walking near it).

“That is to say, I then and there felt the galvanic circuit of thought close; and the sparks which fell from it were the fundamental equations between i, j, k; exactly such as I have used them ever since.”

If you think this is not fun, please, just ignore it. While I’ll write this like talking to a 14-year-old teen, the following is nerdy (mathematical) and lengthy 😅

Today a hundred and four score years ago, Hamilton discovered “quaternions”. To commemorate this, allow me to use (Monero-flavored) quaternions to prove Euler’s identity: If N is a sum of four squares and n is a sum of four squares too, then Nn is also a sum of four squares.

Example: 8 = 2² + 2² + 0² + 0² and 127 = 9² + 6² + 3² + 1² are sums of four squares. So 8*127 = 1016 must be somehow a sum of four squares too.

Proof: Given N = A² + B² + C² + D² and n = a² + b² + c² + d² with some intergers A, B, C, D, a, b, c, d, we need to show Nn = E² + F² + G² + H² with some integers E, F, G, H. Since we’re Monero fans, let us use X, M, R instead of Hamilton’s i, j, k. Things work in a “cyclic“ way like this:

X² = M² = R² = −1 ... Eq.(1)

XM = R, but MX = −R ... Eq.(2)

MR = X, but RM = −X ... Eq.(3)

RX = M, but XR = −M ... Eq.(4)

If we define XMR = −1 imitating Hamilton’s ijk = −1, (2)(3)(4) follow. X, M, R are a bit unusual: the order of multiplication matters (e.g. XM and MX are different). On the other hand, regular numbers (say: e, f, g, h) can “move” freely, as in hXM = XhM = XMh. A quaternion is a “number” of the form e + fX + gM + hR.

Assume we have two quaternions, Q = A + BX + CM + DR and q = a + bX + cM + dR. Multiply Q by q, and things become a bit messy:

Qq = (A + BX + CM + DR)(a + bX + cM + dR)

= Aa + Ab(X) + Ac(M) + Ad(R)

 + Ba(X) + Bb(X²) + Bc(XM) + Bd(XR)

 + Ca(M) + Cb(MX) + Cc(M²) + Cd(MR)

 + Da(R) + Db(RX) + Dc(RM) + Dd(R²)

= Aa + Ab(X) + Ac(M) + Ad(R)

 + Ba(X) + Bb(−1) + Bc(R) + Bd(−M) ← using (1)(2)(4)

 + Ca(M) + Cb(−R) + Cc(−1) + Cd(X) ← using (2)(1)(3)

 + Da(R) + Db(M) + Dc(−X) + Dd(−1) ← using (4)(3)(1)

= (Aa − Bb − Cc − Dd)

 + (Ab + Ba + Cd − Dc)X

 + (Ac − Bd + Ca + Db)M

 + (Ad + Bc − Cb + Da)R

If we write

E = Aa − Bb − Cc − Dd,

F = Ab + Ba + Cd − Dc,

G = Ac − Bd + Ca + Db,

H = Ad + Bc − Cb + Da,

then above mess becomes tidy:

Qq = E + FX + GM + HR ... Eq.(5)

Now, consider a function swap() that converts a given quaternion u = e + fX + gM + hR into a quaternion e − fX − gM − hR. By messy calculation like above, you can show: swap(Q) * swap(q) = E − FX − GM − HR which is = swap(Qq) according (5). Generally, for any two quaternions u, v:

swap(uv) = swap(v) * swap(u) ... Eq.(6)

We define the hash of u = e + fX + gM + hR as hash(u) = e² + f² + g² + h². Since e, f, g, h are regular numbers, a hash is a regular number. Just like above, do some math and you get:

hash(u) = u * swap(u) ... Eq.(7)

Using (7) with u = Qq,

hash(Qq) = (Qq) * swap(Qq) = Q * q * (swap(q) * swap(Q)) ← using (6) with u=Q, v=q

= Q * (q * swap(q)) * swap(Q) = Q * hash(q) * swap(Q) ← using (7)

= Q * swap(Q) * hash(q) ← hash is a regular number; can “move” freely

Again using (7), we conclude hash(Qq) = hash(Q) * hash(q) ... Eq.(8)

Recall the definition of “hash”. Given Q = A + BX + CM + DR and q = a + bX + cM + dR,

hash(Q) * hash(q) = (A² + B² + C² + D²)(a² + b² + c² + d²) ... Eq.(9)

We know Qq = E + FX + GM + HR as in (5), so

hash(Qq) = E² + F² + G² + H² ... Eq.(10)

(8) says (9) = (10), meaning

(A² + B² + C² + D²)(a² + b² + c² + d²) = E² + F² + G² + H² as required.

Example (cont.): With 8 = 2² + 2² + 0² + 0² and 127 = 9² + 6² + 3² + 1²,

E = Aa − Bb − Cc − Dd = 2×9 − 2×6 − 0×3 − 0×1 = 6

F = Ab + Ba + Cd − Dc = 2×6 + 2×9 + 0×1 − 0×3 = 30

G = Ac − Bd + Ca + Db = 2×3 − 2×1 + 0×9 + 0×6 = 4

H = Ad + Bc − Cb + Da = 2×1 + 2×3 − 0×6 + 0×9 = 8

Sure enough, 6² + 30² + 4² + 8² = 1016 = 8*127 😃

Notes: We implicitly assumed that multiplication of quaternions is associative. This assumption is correct as you can see (ij)k = (k)k = −1 and i(jk) = i(i) = −1 are identical, etc. Euler originally used −B, −C, −D, instead of our B, C, D. Both versions are essentially the same.

Monero-themed names ~ Standard names:

X, M, R ~ i, j, k

swap ~ conjugate

hash ~ norm (or norm squared, depending on how you define it)

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 2 points 11 months ago (1 children)

Thanks for nice words! This is like a puzzle. Numbers like 0² = 0×0 = 0, 1² = 1×1 = 1, 2² = 2×2 = 4, 3² = 3×3 = 9 are called squares: 0, 1, 4, 9, 16, 25, 36, 49, … are squares.

Puzzle 1: Can you make 5 by adding two squares? Easy: 2² + 1².

Puzzle 2: How about 73? A bit harder but, yes, 8² + 3² = 64 + 9 = 73.

Knowing these two recipes—how to make 5 (or 73) as a sum of two squares—, there is a way to make 5×73 = 365 as a sum of two squares too. The above is similar, except it’s about a sum of four squares, not two.

A prime like 5, 13, 17 (bigger by 1 than a multiple of 4) can be written as a sum of two squares. On the other hand, a prime like 3, 7, 11 (bigger by 3 than a multiple of 4) can’t be written as a sum of two squares. This looks like a problem of natural numbers, but a natural way to understand what’s going on here is, unexpectedly, seeing this as a problem of the world of complex numbers. That’s counter-intuitive, but fascinating…

What if you’re allowed to use three squares? You can now make 3 = 1² + 1² + 1² and 11 = 3² + 1² + 1². But you can’t still write 7 as a sum of three squares. Then, what if you’re allowed to use four squares? That’s the neat part: you can write any natural number as a sum of four squares, like 7 = 2² + 1² + 1² + 1². A natural way to see what’s going on here is, using quaternions!

If you know a complex number like C = A + Bi and c = a + bi, you can calculate Cc = (A + Bi)(a + bi) like normally, except whenever you get i², you’ll replace it with −1. Quaternions are similar. You can add, multiply, etc. them like normally, except whenever you get i², j², k², ij, ji, jk, kj, ki, ik, you’ll replace them in a certain way (in the post, X, M, R are used instead of i, j, k for fun).

Seriously, math of the above post is not as hard as you may be thinking. The “joke” (not math) may be a bit esoteric for regular people, though, as it’s related to the nature of Monero (“fungible”).

[–] BlackXanthus 2 points 11 months ago (1 children)

Comrade lemming, I love your enthusiasm. I think I get the basic idea that a quaternion is a number made up of 4 squares.

I love your belief that I know a complex number! (I don't).

I would love to have a better grip on maths, but it has always alluded me. I once got a maths PhD student to tutor me, and they said I lacked a fundamental understanding of number.

I will keep reading, it may click!

Thanks for your reply!

[–] [email protected] 1 points 11 months ago

Yeah, something like that. A quaternion has 4 elements, and its “size” is related to the sum of 4 squares.

I’m not that good myself, just yet another number theory lover (or a wannabe Gauss, a wannabe Knuth…) You at least knew the word quaternion and more or less understood immediately that it’s related to a (four-dimensional) vector; that’s 100% correct & you do have a good grip, or good intuition. Anyway, the original post is basically a joke, where something rather trivial is written in an unusual way. Not to be taken too seriously…