Tuesday, December 27, 2022

Erdős–Straus Conjecture

 Consider a circle of perimeter length \(4\),


we divide this circle into \(n\) sector.  We divide one of these sectors of angle \(\cfrac{2\pi}{n}\) further into \(M\) sectors such that,

\(M=4*\cfrac{N}{n}\), 

\(M\ge 3\), both \(M\) and \(N\) are integer

This way,

\(\cfrac{1}{M}*\cfrac{2\pi}{n}*N=\cfrac{n}{4N}\cfrac{2\pi}{n}*N=\cfrac{\pi}{2}\)

That is we can walk \(N\) step of \((\cfrac{1}{M}*\cfrac{2\pi}{n})\) to \(\cfrac{\pi}{2}\) along the circumference for a total distance of \(1\).


On the sector with \(M\) markings, we divide it with two radial lines drawn through the markings into three sectors.  Obviously,

\(a+b+c=M\)

If \(a\) is on the first marking, \(N\) steps will take it  down \((\cfrac{\pi}{2})\) to \(1\), ie. the length of \(a\)

\(\bar{a}=\cfrac{1}{N}\)

Then the next line divides the remaining \(M-1\) into \(b\) and \(c\), such that both \(b\) and \(c\) divides \(N\) (The proper choice of \(M\) and \(N\) will make this possible and easy).  We formulate,

\(b.B=N\) and \(b.C=N\)

then in \(B\) steps \(b\) will travel through \(\cfrac{\pi}{2}\) for a distance of \(1\) along the circumference.  Similarly, in \(C\) steps \(c\) will travel through \(\cfrac{\pi}{2}\) for a distance of \(1\) along the circumference.  That is to say,

\(\bar{b}=\cfrac{1}{B}\)   and 

\(\bar{c}=\cfrac{1}{C}\)

Since, the size of this sector is \(\cfrac{4}{n}\), we have

\(\bar{a}+\bar{b}+\bar{c}=\cfrac{1}{N}+\cfrac{1}{B}+\cfrac{1}{C}=\cfrac{4}{n}\) for any \(n\ge 2\) and \(N\), \(B\), \(C\)

all integer.  Provided \(B\) and \(C\) can be found.

For the proper choice of \(M\) and \(N\);

When \(n\) is even we chose \(N=n\) then \(M=4\).

 \(a\) is found as above, one marking wide, and takes \(N\) step to reach \(1\) along the circumference

\(\bar{a}=\cfrac{1}{N}\)

\(b\) is set to \(2\) markings wide such that \(b\) will take \(\cfrac{N}{2}\) steps to travel a distance of \(1\) so,

\(\bar{b}=\cfrac{2}{N}\)

since \(N=n\) is even, \(\cfrac{N}{2}\) is an integer.

After taking step \(a\), one step,

\(b+c=M-1=4-1=3\)

After taking step \(b\), two steps,

\(2+c=3\)

then only one step \(c=1\) remains and so,

\(\bar{c}=\cfrac{1}{N}\)

since \(a\), \(b\) and \(c\) make up the sector of length \(\cfrac{4}{n}\)

\(\cfrac{1}{N}+\cfrac{2}{N}+\cfrac{1}{N}=\cfrac{4}{n}\)

we have proven Erdős–Straus conjecture for even numbers.

When \(n\) is odd, it has at least one odd factor \(f_o\).  We let

\(n=f_of_1\)  where \(f_1\) is another odd factor and may be one.

 Let

\(N=\cfrac{f_1}{2}(n+f_o)(f_o-1)=\cfrac{f_1}{2}(f_1f_o+f_o)(f_o-1)\)

\(N=\cfrac{f_of_1(f_1+1)(f_o-1)}{2}=\cfrac{n}{2}(f_1+1)(f_o-1)\) 

since \((f_1+1)\) \((f_o-1)\) are even, \(N\) is an integer.

So, \(M=4\cfrac{N}{n}=\cfrac{2n(f_1+1)(f_o-1)}{n}=2(f_1+1)(f_o-1)\)

We let

\(a=f_1(f_o-1)\)

the first step we take is \(f_1(f_o-1)\)

\(\bar{a}=\cfrac{1}{K}\)

\(K=\cfrac{f_of_1(f_1+1)(f_o-1)}{2f_1(f_o-1)}\) since \((f_o-1)\) and \((f_1+1)\) are even, \(K\) is an integer.

Let,

\(b=f_o-1\)     \(\bar{b}=\cfrac{1}{J}\)

\(J=\cfrac{f_of_1(f_1+1)(f_o-1)}{2(f_o-1)}\) \(J\) is an integer. 

and the length of this step is \(\bar{b}=\cfrac{1}{J}\)

After step \(b\),

\(c=2(f_1+1)(f_o-1)-f_1(f_o-1)-(f_o-1)\)

\(c=2f_of_1+2f_o-2f_1-2-f_of_1+f_1-f_o+1=f_of_1+f_o-f_1-1\) 

\(c=f_1(f_o-1)+(f_o-1)=(f_o-1)(f_1+1)\) 

But we take half step instead,

\(c=\cfrac{(f_o-1)(f_1+1)}{2}\) 

since \((f_o-1)\) and \((f_1+1)\) are even, \(c\) is an integer.

\(\bar{c}=\cfrac{1}{L}\)

\(L=\cfrac{2f_of_1(f_1+1)(f_o-1)}{2(f_o-1)(f_1+1)}\)

and \(L\) is an integer.

The total length of these steps \(a\), \(b\) and \(c\) is \(\cfrac{4}{n}\), we have,

\(\cfrac{1}{K}+\cfrac{1}{J}+\cfrac{1}{L}=\cfrac{4}{n}\)

where \(K\), \(J\) and \(L\) are integer.

and we have proven Erdős–Straus Conjecture for all \(n\), \(n\ge 2\)

Good morning.

Notes:

The steps, \(a\), \(b\) and \(c\) are sectors in units of markings provided by dividing the sector \(\cfrac{4}{n}\) by \(M\).  These steps size are chosen such that,

\(a\).\(A\)=\(N\),     \(b\).\(B\)=\(N\) and     \(c\).\(C\)=\(N\)

Given step size \(a\), it will take \(A\) step to reach \(N\),  where \(\cfrac{\pi}{2}\) is through a lenght of \(1\).  So, the length of \(a\) is,

\(\bar{a}=\cfrac{1}{A}\)

similarly for \(b\) and \(c\).  \(M\) goes down to \(\cfrac{4}{n}\), \(N\) goes down to \(1\).