ITT we prove the Erdős–Straus conjecture.

>>8111821
have you tried induction?

>>8111968
Are you a computer scientist?

>>8111821
n also element of N?

>>8111972
Yes.

>>8111821
[math]\frac{4}{3} = \frac{1}{2} + \frac{1}{2} + \frac{1}{3}[/math]

>>8111821
[math]\frac{4}{4} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3}[/math]

>>8111821
[math]\frac{4}{5} = \frac{1}{2} + \frac{1}{5} + \frac{1}{10}[/math]

>>8112085
Yeah, OP goofed that one, it's n >= 2, not n > 2.

>>8111968
Induction is only useful when you can write the n+1 case in terms of the n case, which I don't think you can do here because n is a part of the denominator.

First let [math]n = 4a + b[/math]. Whenever [math]b = 0[/math] then [math]\frac{4}{n} = \frac{1}{3a} + \frac{1}{3a} + \frac{1}{3a}[/math]. So for a quarter of cases there exist [math]x, y, z[/math].

>>8112097
Therefore, the conjecture is true QED.

>>8112085
>>8112087
These aren't proofs, just examples.
You need to find a way to show that EVERY n > 2 fits the conjecture.

>>8112097
Next when [math]b = 1[/math].

If [math]a = 1[/math] mod 2 then

[math]\frac{4}{n} = \frac{1}{a+1} + \frac{1}{n \ceil{a/2} + \frac{1}{2 n \ceil{a/2}

otherwise it's something else. That's all I have.

>>8112103
[math]\frac{4}{n} = \frac{1}{a+1} + \frac{1}{n \ceil{a/2}} + \frac{1}{2 n \ceil{a/2}}[/math]

[math] \frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \implies \frac{n}{4} = x+y+z [/math] which can only be true for even [math] n [/math] thus the conjecture is false for trivial reasons.

case closed

>>8112105
Nice counterexample

>>8112092
Hmm... so [math]\frac{4}{5} = \frac{1}{2} + \frac{1}{5} + \frac{1}{10}[/math], but then if we take [math]2 = 2_3[/math] and [math]5 = 12_3[/math] and [math]10 = 101_3[/math], so then if we take [math]1(1^{2}) + 1(1^{1}) + 6(6^{0})[/math], where each outside factor and base under the exponent corresponds to the corresponding digit position in the sum of those three numbers in their base 3 form, we then get 3 as a result, which is the number of values which it took to form the number \frac{4}{5} as a sum of reciprocals. Coincidence?

>>8112105
that's not how reciprocals work.
n = 1/2 + 1/3 =/=> 1/n = 2 + 3
n = 5/6 =/=> n = 5

>>8112100
It's called proof by exhaustion, bitch.

>>8111821
Well, what if we start by creating a topology on the space of solutions where closeness has to do with the factorization of n and the number of solutions for x, y, and z that work. That way we can just look at the topological space and see where things collapse to a 0 because there isn't an x or a y or a z to use to express that point in the topological space, and we could approach this by creating a limit method where you take a limit in this space and show that the limit does not go to 0, because if you can show that the limit does not go to 0, then you have shown that there can be an x and a y and a z that solve it at that point so then you just do this at all of the points, and then you've shown that there is always an x and a y and z for that given n at all of the points because those points correspond to n values. Now, I'm thinking we should make the open sets something like tuples like (x-n, y-z, n+z), that way it's in a compressed form, but we can still recover all of the values through linear combinations.

>>8111821
[math]\frac{4}{6} = \frac{1}{4} + \frac{1}{4} + \frac{1}{6}[/math]

>>8111821
[math]\frac{4}{7} = \frac{1}{4} + \frac{1}{4} + \frac{1}{14}[/math]

>>8111821
[math]\frac{4}{8} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6}[/math]

>>8112095
>Induction is only useful when you can write the n+1 case in terms of the n case, which I don't think you can do here because n is a part of the denominator.

oh I'm laffin

File: Guardian.jpg (60KB, 800x567px) Image search: [Google]

60KB, 800x567px

>>8112154
>>8112155
And so on and so on.

qed

>>8111821
Assume for some given n, that a solution using x, y, and z exists. That is, [math]\frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}[/math]. Then we can also form a solution for any value m such that n divides m.

Since n divides m, we have that [math]m = kn[/math] for some natural number k. Then [math]\frac{4}{m} = \frac{4}{kn} = \frac{1}{kx} + \frac{1}{ky} + \frac{1}{kz}[/math].

Therefore, we can reduce this down to solving for all prime n.

>>8112213
Wouldn't that be infinity? Aren't you just summing 4/n for all n > 2?

>>8112218
im summing up 1/n which leads to e, or rather from 2 on, so its e-1. im no math, hoped this could help

>>8111821
just brute force it

>>8112232
already done for 1<n<10^14.

>>8112184
Well, can't you just brute force it by this point?
It's not like there are infinitely many prime numbers.

>>8112236
Yes there are.

>>8112234
That's enough to assume that it's true.

File: 52.jpg (65KB, 500x499px) Image search: [Google]

65KB, 500x499px

>>8112246
therefore its called a conjecture.

>>8112105
>case for your knowledge of fractions closed
FTFY

>>8112100
then you'd better get started now.

4/n = (xy + yz + xz)/xyz

There you go friends I rewrote the equation for you. I'm going to go take a nap, the rest is pretty simple. If you need help, I left an ingenious explanation on a sheet of paper somewhere around here, for there wasn't enough room in this post.

>>8112234
What about n=10^14?

>>8112332
x=5*10^13
y=1*10^14
z=1*10^14

>>8112329
8/n is equal to the ratio of the surface area of a rectangular prism of sides x,y,z to its volume.
Neat

>>8111821
Well at least 2 have to be even.
since
xy + yz + zx = 4k
k being a reducible factor that is shared with xyz
so the only way for this to be true is either
odd + even + odd which is impossible since all 3 terms are a produce of 2 variables that appear twice, so there can't exist only 1 even term
or
even+even+even
which would mean that either 2 variables are even or 3 of them are even.

>>8112329
>>8112340
I don't know if it's of any help, but almost all numbers can be written as xy+yz+xz for some x,y,z.
The ones that don't are called idoneal numbers. We know 65 of them, and it's very unlikely that there's another one.

>>8112095
>>Induction is only useful when you can write the n+1 case in terms of the n case, which I don't think you can do here because n is a part of the denominator.
brainlet

>anonymous shitposter solves old conjecture in number theory, claims it was "easy"
>thousands of mathematicians butthurt to such an extent that number theory research is effectively halted for 7 years

make it so

>>8112344
>We know 65 of them, and it's very unlikely that there's another one.
In 1973, Peter J. Weinberger proved that at most one other idoneal number exists, and that the list above is complete if the generalized Riemann hypothesis holds.[3]

>>8112364
ye

>>8112146
I don't know topology but try it show us what you get

>>8111821
[math]\frac{4}{9} = \frac{1}{3} + \frac{1}{18} + \frac{1}{18}[/math]

>>8111821
[math]\frac{4}{10} = \frac{1}{5} + \frac{1}{10} + \frac{1}{10}[/math]

>>8111821
[math]\frac{4}{11} = \frac{1}{3} + \frac{1}{66} + \frac{1}{66}[/math]

>>8113694
>>8113685
>>8113684
>>8112155
>>8112154
>>8112148
>>8112092
>>8112087
>>8112085
please see>>8112184

this hasn't had a proof written for it yet?

>>8113724
https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture
There's an "unsolved" thingy next to it

>>8111821
It's nice that it's such an easy-to-understand unsolved conjecture, but is there any reason we should care in terms of relationships to other area of number theory?