Watch this: [math](x + y)^p = x^p + y^p[/math]

At first, I asked myself why I had never seen that property before.
Then I realised it's just
F E R M A T ' S L I T T L E T H E O R E M

Disappointing.

If your aren't a brainlet who's had his brain rotted by anime you'd realize it's much simpler to simply prove this result and then take Fermat's Little Theorem as an instant corollary.

>corollary
more like a lemma you pleb

I mean, binomial expantion gives the problem away, but how does the thesis yield that [math](x + y)^p = x^p + y^p[/math] is actually [math]x+y[/math]?

Idk, Fermat's Little Theorem is already as easy as it gets to prove, also its proof is instructive on how you can use inverses modulo p

[math](4 + 3)^3 = 343[/math]
[math]4^3 + 3^3 = 91[/math]

>3
>prime

cause [eqn]x^p =(x-1)^p + 1^p =(x-1)^p +1[/eqn] do this a few more times and you'll have x many 1s there.

>3 isn't prime

I see. Well, that's a nice proof, too. Surely better than that meme necklace one

idk just something I figured out a while backs I never learned the necklace one cause I'm too stupid lol

because [math]a = \sum_{1}^{a}1[/math]

hello op i must inform u this is totally not correct i will show u real way first of all u take the x and make it so that if u put x multiply by x then u have 2x or x2 and then u multiply all 2 and u have x2y or xy2 short and then y is 2x too so result is x2 * xy2 * y2

Now divide both by 3 and take the remainders

What necklace?

>1 + 0 = 1
Wow, you discovered one of the basic axioms of mathematics! When will you get your fields medal?

Watch thia x^i=-x wao much euler very smurt

the first 5 primes are 2 3 5 7 11, brainlet

CONGRUENCE IS NOT EQUALITY

wtf

isn't this just basic algebra?
what's the joke

What are the factors of 3 then huh
ya fggt get fucking rekt im much smarter than you
hahaha brainlet hahaha

It is if you're in the right ring

Oh this can be an interesting question: What's the lower bound of the error when you use the approximation :
[eqn] (x+y)^p \; \approx \; x^p \, + \,y^p [/eqn]
Also what condtitions can be imposed on [math]x,y[/math] and [math]p[/math] to have a relative error level under a certain threshold [math]\epsilon[/math] i.e. conditions such as
[eqn]\left|\frac{x^p+y^p}{(x+y)^p}\right|

shut up

Seems you've failed math101

[eqn]\left|\frac{x^p+y^p}{(x+y)^p}\right|

Wouldn't you want
[eqn]\left|\frac{x^p+y^p}{(x+y)^p}-1\right|

see OP wrote exactly what he wanted, the point was to get surprised reactions from people who don't know math and don't instantly recognize what he posted as true

>1 < eps for any eps
idiot

>missing the point
>dividing by a non-unit
>comparing its absolute value and not its distance to 1
brainlet

I thought you couldn't distribute exponents to added terms?

guess that's correct, if x and y are orthogonal vectors and we're talking about taking the inner product.

you can if the field has characteristic [math]p[\math]

>If I leave away important information people will think it's not a tautology, even though it is when I assume [math] x,y \in \mathbb{Z} ~,~ p ~\text{is prime}[/math] and it's actually all mod p.
>LeL brainlets

2*1

Let [math] x=1, y=3, p=2 [/math]

>Use non-standard notation
>Make false statement that is only true in the particular ring I was thinking about
Haha tricked you brainlets can't read my mind

k

>taking the pth power is non standard notation
b r a i n l e t

if you didn't see and use this identity a lot in your algebra sequence, you shouldn't post in math threads

If p = 0 it would be 1 = 2.

I won't be mean to you. p is 0 mod p.

That's not what he was referring to by nonstandard notation, moron.

its such a common notion that it has its own name.
>en.wikipedia.org/wiki/Frobenius_endomorphism

Fermat BTFO

you forgot 1 number...

>1 is prime
I want brainlets to leave

1 is a super prime, mathematicians are just too lazy to acknowledge it

If [math]p[/math] is a prime number and [math](x,y) \in ( \frac{ \mathbf{Z} }{ p\mathbf{Z} } )^2 [/math] it works. Using Newton's binomial formula, it is obvious.