ITT we prove the Erdős–Straus conjecture

Therefore, the conjecture is true QED.

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

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.

[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

Nice counterexample

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?

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

It's called proof by exhaustion, bitch.

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.

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

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

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

>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

And so on and so on.

qed

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.

...

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

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

just brute force it

already done for 1

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

Yes there are.

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

therefore its called a conjecture.

>case for your knowledge of fractions closed
FTFY

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.

What about n=10^14?

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

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

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.

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.

>>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

>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]

ye

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

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

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

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

please see

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

en.wikipedia.org/wiki/Erdős–Straus_conjecture
There's an "unsolved" thingy next to it

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?