Why tho? Not gonna lie, brainlet here tho who took one term

by the binomial theorem
(m+r)^x
= sum_{i=0}^x (x choose i) m^i r^{x-i}

each summand with 1<= i <= x is divisible by m so is equal to 0 mod m

so
(m+r)^x
= sum_{i=0}^x (x choose i) m^i r^{x-i}
= r^x (mod m)

>>9137794
Triple induction would do it.

>>9137794
You can literally expand the brackets (obviously you don't need to write out every term), and you'll see you have a bunch of terms of the form m*something with a lone r^x at the end, which is clearly congruent to r^x (mod m).

>>9137804
huh. interesting. Thank you anon.

>>9137806
Try quadruple brainlet.

File: proof.png (44KB, 515x590px) Image search: [Google]

44KB, 515x590px

>>9137804
thanks anon.

it worked flawlessly

>>9139654
That was painful to read. You need to start with the left side of that congruence relation ((m+r)^n), then show that it is congruent to the right (r^n mod m). The way you have written it is confusing and not logically sound.

>>9139673
Do I?
but if I turn the top m | (m+r)^n -r^n into its modular form it becomes (m+r)^n - r^n = 0 (mod m)

what is the reason to skip a step and start with r^n on the right side of the congruence?

>>9139677
You can start with (m+r)^n - r^n = 0 (mod m) if you want. The point is you're proving one side is congruent to the other. Your proof can't start by writing the statement you're trying to prove.
So, for instance, you might start with (m+r)^n then show that (m+r)^n = ... = m*(something) + r^n = r^n (mod m). Then the rest is obvious.

>>9139682
>Your proof can't start by writing the statement you're trying to prove.
huh.
so I actually came up with this theorum when asked to do an induction proof on 4 | 5^x -1 a few years ago.
would it be proper to start with 4 | 5^x -1 and prove it by the proof above, since 4 | 5^x -1 is the same as m=4, r=1, i.e. 4 | (4+1)^x - 1^x ?

>>9139687
Well I wouldn't use the proof you posted because it's nonsense. But sure, 4 | 5^x - 1 does follow from m|(m + r)^n - r^n.

>>9139687
>>9139695
Also, do you not understand why starting a proof by assuming the statement you're trying to prove is ridiculous?

>>9139698
Not fully.

If I want to prove 5+3 = 8 why can I not start with 5+3 =8 and either subtract 3 from both sides or add 5+3. Both times im starting with the statement im trying to prove and in one instance I operate on both sides of the equation

>>9139705
If you start by assuming what you're trying to prove then there's nothing left to do, its truth has already been established.

>>9139705
Because it's circular reasoning. I can't just prove unicorns exist by first assuming unicorns exist and saying "wow, look at that, they exist". How the fuck are you familiar with what an induction proof is, yet you don't understand the absolute basics of logical reasoning?

>>9139712
>>9139713
ive taught myself my own math, i dont have a formal education in it.

I buy books and I read them and mess around with numbers. Thats how I picked up induction, modular arithmetic and had anon mention binomial expansion up here >>9137804 and so I read about it today.

I have a lack of formal math education which probably would have taught me the basics of proofs.

I know this sounds stupid, but why do I have to abstract the problem into something else?

why can you not assume its true? isnt it the opposite of a proof by contradiction where I just assume its false, and when the math shows me it cant be false, then it must be true?

I start out assuming that it is true, and if the math shows that it is true, then it must be true. If I start out assuming its true and the math doesnt work out, then it may not be true.

Why can you not do this?

>>9139717
>ive taught myself my own math
I don't mean to sound harsh, but you've done a terrible job of it.
Please read something like 'How to Prove it' by Velleman.

>>9139723
ill do that

>>9139717
>>9139723
>>9139724

The thing with mathematics is since it is so stringent/strict/rigorous it is, of all forms of logic, the most 'not wanting to reach an answer from self, only consensus'.

However maths is just you and numbers, so the consensus is formed through whatever axiom/rule/law has been set out.

Maths rejects itself to be measured, so it can retain its identity as 'measurer'.

So, yes. You are a mathematician and actually pretty clever given what you've just resolved (It's a 3-sat variant if you're curious).

>>9137794
>tho
Stop doing that