This is a partial proof for the famous odd perfect number conjecture .
Consider any number N = pa1 * pa2 * ... pak . ( Note a1,...ak are the indices for prime numbers , Actually N can be of the form pa1^r1 * pa2^ r2 *.... pak^rk . However i am interested only in those numbers which can be expressed as N = pa1 * pa2 * ... pak ) . Thus the factors would be 1,pa1 , pa2 , .. pak , pa1*pa2,.. pa1*pak, pa2*pa3 ,.. Similarly taken three at a time and so on till taken all at a time.
A number is perfect if all the factors expect the one taken all at a time i.e N ( or all the prime factors taken all at a time ) should be equal to N.
If the above statement is not clear then let me explain it with a number which has just three factors i.e. N =pa*pb*pc.
Then N is perfect if 1+ pa+pb+pc+ pa*pb + pb*pc + pa*pc = pa*pb*pc.
i.e 1/ pa*pb*pc + 1/ pb*pc + 1/ pa*pc + 1/pa*pb + 1/pa + 1/pb + 1/pc =1.
i.e (pa+1)* (pb+1) * ( pc +1) / pa *pb*pc =2 . ( This is just rearrangement of the above equation).
The similar thing can be expanded for k prime factors also.
Thus we have (pa1 + 1)* ( pa2 +1) *..... *( pak +1) / pa1*pa2*... *pak = 2.
I have written pa1, pa2 ,.. pak in ascending order .
Thus if N is odd then none of the pak's = 2.
Also all the number in the numerator can be can be expressed as 2^k * pbk . It is trivial that pbk < pak.
(Note if pa1 is 3 . then it becomes 2^2 * 1 . No pbk term this is an exception).
We note that there is no term in the numerator which will cancel pak( largest prime number) . Thus the ratio is never an integer thus it is not equal to 2 also . Hence there doesn't exist any odd perfect number N = pa1 * pa2 * ... pak .