/sci/ - Science & Math

shit that's like, real math and stuff

>>2971382

Any tips on how to learn about eliptic curves/modular forms? Books or anything?

>>2971382

What is the field F_7?

>>2971382
Sorry bro, that's not calc 1, don't fucking post in it /sci/.

>>2971382

Do your own homework, lazy stupid first year graduate student.

>>2971397

This chump doesn't know what he's talking about. What does the olympiad have to do with anything? He's just trolling and/or trying to make himself feel smart because it has no solutions.

(x1,x2) = (0,-5) of course! This is simple vector arithmetic. LOL, how can you not understand this?

XD

>>2971397
I can't suggest any books in English because when I studied it we mostly relied on lecture notes in our own language but the Wikipedia page on elliptic curves looks like a decent starting point. No joke.
>>2971412
<span class="math">\mathbb{F}_7 = {0, 1, \dots, 6}[/spoiler]

>>2971453

25 ~ 0 != 3

>>2971438
>What does the olympiad have to do with anything?
Challenging problems.
>it has no solutions
It has. You'd know that if you knew the subject at hand.

x1 = 0, x2 = -1

(1)

f(0)^2 = 0^3 + 4*0 + 3
f(0)^2 = 3
f(0) = sqrt(3)

(2)

f(-1)^2 = -1^3 + 4*(-1) + 3
f(-1)^2 = -3 - 4 + 3
f(-1) = sqrt(-4)
f(-1) = 2i

Q.E.D.

Please comment.

What are we exactly trying to find here?

>>2971494

> implying sqrt(-4) = 2i

>>2971382

I think i know what has to be done here, but i do not know an analytical way.

I guess i can start with th (1,1) and write it as (6k+1,6k+1), then substract the (1,6) and get (6k,6k-5) or (6k,6k+1).

Is (6k,6k+1) the general solution for all natural numbers k?

Apparently my university has one of the top minds in the study of elliptic curve cryptography, so I went to a lecture of hers to see what it was all about.

My jaw was on the floor the whole time.

I've almost finished by B.S. in mathematics and that shit was wayyyy over my head.

>>2971505

I mean, as an example for k=1, (6,7) would yield

49 = 216 + 24 + 3 = 243

Where 49 ~ 1 and 243 ~ 3

and i see it didn't work out, shit.

>>x2 = -1
>>MFW this is what I was actually thinking.

>>2971516
>>2971505
Look that the OP picture. It is a hint as to what you have to do.

>>2971453
If you're serious, I would seriously consider jumping off a bridge if I were you.

>>2971516

god i'm dumb.

(7k+1,7k+1) - (1,6) = (7k,7k+2)

for k = 1: (7,9)

Testan: 81 = 343 + 28 + 3 = 374

With 81 ~ 4 and 374 ~ 3

and it still didn't work, shit.

>>2971524

hm

(x_1,x_2) + (1,6) - (1,1) = 0 ?

>>2971546
Right. Now read up on en.wikipedia.org/wiki/Elliptic_curve#The_group_law

>>2971558

Wish i was firm in group theory. Ok, what i get now is that

P = (1,6)
Q = (-1,-1)

s = 7/2

And though i'm not sure right now why this makes sense, the third point is

x_1 = 49/4
x_2 = 363/8

according to the definition on wikipedia. Or will i have to take everything mod 7 so that s = 0?

In that case

x_1 = 0 (or 7k)
x_2 = 6 (or 7k-1)

Trying this for k=1 so that R = (7,6):

36 = 343 + 28 + 3 = 374

36 ~ 1 and 374 ~ 3

Hm, seems my brain took a weekend off or this is too high for me.

>>2971616

wait wait wait what the fuck was i doing?

in case s = 0 we have

(7k,7k+6)

so for k = 1: (7,13)

49 = 2212

0 = 0

Did i get it finally?

>>2971632

No, i didn't. Shit.

>>2971616
>>2971655
>>2971632
First of all, -(1,1) is not the same as (-1,-1) = (6, 6) in this group due to the way "+" is defined. For one thing, (6, 6) doesn't satisfy the curve's equation. Instead you can assume that <span class="math">(x_1, x_2)[/spoiler] is P, (1,6) is Q and (1, 1) is R and use the same formulas you're using now to produce the (rather complicated) equations to solve for <span class="math">(x_1, x_2)[/spoiler]... Or you can find all the points from <span class="math">\mathbb{F}_7[/spoiler] that lie on the curve and, well, chеck 'em one by one.

Okay, do you want the answer?

>>2971714

Yes, that would be great. I was struggling around a bit but can't wrap my head around it right now.

Thanks for your time, kind sir.

>>2971734
The answer is (5,1). To be able to solve these more easily I suggest that you brush up on modular arithmetic and finite fields. Reading http://www.springeronline.com/sgw/cda/pageitems/document/cda_downloaddocument/0,11996,0-0-45
-110359-0,00.pdf might help.

>>2971762

Great, thank you!

>>2971762
You know what? No more problems for today but I'll be back with something easier tomorrow. Thanks to everyone who tried to solve this one.

>>2971776
You're welcome. Come back tomorrow for more.

>>2971762
I keep getting (5,6).
To start, (1,6) = -(1,1) right?
To solve
(1,6) + Q + 0 = 0
we would reflect about the x axis (sub -y for y)?
So that we need to solve
<span class="math">(x_1, x_2) + (1,6) + (1,6) = 0[/spoiler]
?
Then I get a slope of 0, which leads me to solve the curve equation for the other possible x for y=6, giving (5,6).
I don't know anything about elliptic curves besides what I read on Wikipedia right now, so tell me if I'm doing something wrong.

bump

>>2971868
(1,6) + (1,1) = (5,6) <span class="math">\neq[/spoiler] 0 (i.e., the identity element). Therefore according to the definition of a group (1,6) isn't the inverse of (1,1).

>>2972017
><span class="math">(1,6) + (1,1) = (5,6) \neq 0[/spoiler]
How do you get that?

>>2972035
Never mind, I was wrong.

>>2971868
OP here. Yes, (1,6) = -(1,1).
<span class="math">(x_1,x_2)[/spoiler] = (1,6) - (1,1) = (1,6) + (1,6) = (5,1). Take note of that minus sign in the formula.

>>2972144
Shouldn't that be <span class="math">(x_1,x_2) = (1,1) - (1,6)[/spoiler]?

>>2972154
Fuck, I screwed up. You are right.

<span class="math">(x_1, x_2)[/spoiler] = (1,1) - (1,6) = (1,1) + (1,1) = (5,6).

My initial answer was wrong because apparently it is I who should mind the minus signs. Sorry, everyone.

>>2971762

Hey OP - I know you think you're super smart, but I said there is no solution, and there is none, and if you check your arithmetic (which you are so proud of) you'll notice that (5,1) is not a solution to this equation.

(5,1)+(1,6) = (6,0)

which does not satisfy the equation:

1^2 =/= 5^2 + 4 * 5 + 3

(And yes, I'm doing all my arithmetic mod 7.)

I have a PhD in mathematics, and while I do not study "elliptic curve cryptography" (this is hardly a "branch" of mathematics, it's called algebraic number theory), I know better. Even a computer running Sage, GAP, Maple, etc. can verify this.

>>2974631
erm, see the post right above yours