Ok so here is the starting conjecture
4/n = 1/a + 1/b + 1/c
this has been tested up to 10^17 or something huge like that, however it is still a conjecture no proof of it exists.
I figured out 1/4 of the problem but we’ll need to introduce this a little bit more first.
4/3
4/4*
4/5
4/6
4/7
4/8*
4/9
4/10
4/11
4/12*
4/13
4/14
4/15
4/16*
notice something about those with the stars ?
it is the suite 1/1, 1/2, 1/3, 1/4 disguised.
notice something about the regularity ? it is one in every 4 fractions, those are the only one with remainder 0 after division by 4. 4 is the theme after all it is the numerator.
4/4
4/8
4/12
4/16
…
now let me explain why I say this is 1/4 the conjecture
there is three other parts
-remainder 1- -> form 4n+1
4/5
4/9
4/13
…
-remainder 2- -> form 4n+2
4/6
4/10
4/14
…
-remainder 3- -> form 4n+3
4/7
4/11
4/15
…
now lets explore what the remainder 1 gives us
1/1 = 1/2 + 1/3 + 1/6
1/2 = 1/4 + 1/5 + 1/20
1/3 = 1/6 + 1/7 + 1/42
1/4 = 1/8 + 1/9 + 1/72
…
here it is 1/n = 1/2n + 1/(2n+1) + 1/2n(2n+1)
so this is 1/4 the proof to the Erdős–Straus conjecture, but this is no proof not anymore than the conjecture here we go for the proof via contradiction:
n(1/n) = n(1/2n + 1/(2n+1) + 1/2n(2n+1))
1 = 1/2 + n/(2n+1) + n/2n(2n+1)
1 = 1/2 + n/(2n+1) + n/(4(n^2)+2n)
1/2 = n/(2n+1) + n/(4(n^2)+2n)
1/2 = n/(2n+1) + n/n(4n+2)
1/2 = n/(2n+1) + 1/(4n+2)
let’s take a break to appreciate this
1/2 = 1/3 + 1/6
1/2 = 2/5 + 1/10
1/2 = 3/7 + 1/14
1/2 = 4/9 + 1/18
1/2 = 5/11 + 1/22
1/2 = 7/13 + 1/26
This is at the center to this phenomenon here .
So we have
1/2 = n/(2n+1) + 1/(4n+2)
let’s multiply by 4n+2 to finishe this
(4n+2)(1/2) = (4n+2) (n/(2n+1) + 1/(4n+2))
(4n+2)/2 = 4(n^2)+2/(2n+1) + (4n+2)/(4n+2)
2(2n+1)/2 = (2n+1)(2n)/(2n+1)+1
2n+1 = 2n+1
1 = 1
this means that if we would replace the original statement
1/n = 1/2n + 1/2n+1 + 1/2n(2n+1)
by a contradiction to the original statement
1/n != 1/2n + 1/2n+1 + 1/2n(2n+1)
than we would arrive at the absurd by following the previous line of valid algebraic transformation. Invalid statement flow from invalid statement and valid statement flow from valid statement, since 1 cannot not equal 1 than the statement that leads to the reasoning that 1 would equal 1 would be valid since it cannot be an invalid reasoning since 1 in facts equals one, so it would have to be a valid reasoning and therefore:
1/n = 1/2n + 1/(2n+1) + 1/2n(2n+1)
is a theorem ;).
Now we have 3/4 of the work remaining to the Erdős–Straus conjecture. We did 1/4 of the work the 4n form, we have remaining forms: 4n+1, 4n+2, 4n+3.