Consider a circle of perimeter length \(4\),
\(M=4*\cfrac{N}{n}\),
\(M\ge 3\), 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 \(b\),
\(c=M-a-b\)
\(c=2(f_1+1)(f_o-1)-f_1(f_o-1)-(f_o-1)\)
\(c=2f_of_1+2f_o-2f_1-2-f_of_1+f_1-f_o+1=f_of_1+f_o-f_1-1\)
\(c=f_1(f_o-1)+(f_o-1)=(f_o-1)(f_1+1)\)
But we take half step instead to go further to \(\cfrac{\pi}{2}\) away, when the total length travelled by \(c\) is \(1\).
\(c=\cfrac{(f_o-1)(f_1+1)}{2}\)
since \((f_o-1)\) and \((f_1+1)\) are even, \(c\) is an integer.
\(\bar{c}=\cfrac{2}{L}\)
\(L=\cfrac{2f_of_1(f_1+1)(f_o-1)}{2(f_o-1)(f_1+1)}\)
and \(L\) is an integer.
The total length of these steps \(a\), \(b\) and \(c\) is \(\cfrac{4}{n}\), we have,
\(\cfrac{1}{K}+\cfrac{1}{J}+\cfrac{2}{L}=\cfrac{4}{n}\)
where \(K\), \(J\) and \(L\) are integer.
and we have ChanHL Theorem for all odd \(n\), \(n\gt 2\), that corrects Erdős–Straus Conjecture.
