M and N are amicable. σ(M) is the sum of proper divisors. σ(220)=1+2+4+5+10+11+20+22+44+55+110+220=504.
Moreover, σ(N)=σ(284)=1+2+4+71+142+284=504.
σ(M)-M=N and σ(N)-N=M
In this case, M=220 and N=284.
σ(220)-220=504-220=284
σ(284)-284=504-284=220
∴
σ(M)=M+N=σ(N)
This is crossing over.
M=apq, N=ar
pqr are different prime numbers, and a is the prime relatively.
Leonhard Euler found it.
σ(p)=p+1, p is the prime.
σ(pq)=σ(p)σ(q)=1+p+q+pq=(1+p)+q(1+p)=(1+p)(1+q), p and q are prime numbers.
Therefore,
σ(M)=σ(N), σ(apq)=σ(ar)
σ(a)σ(p)σ(q)=σ(a)σ(r)
σ(p)σ(q)=σ(r)
∴
(p+1)(q+1)=r+1
This is based on prime numbers. Then, you put x=p+1, y=q+1.
xy=r+1, r=xy-1
σ(M)=M+N=apq+ar=a(pq+r)
σ(M)=σ(a)σ(p)σ(q)=σ(a)(p+1)(q+1)=a(pq+r)
σ(a)xy=a[(x-1)(y-1)+(xy-1)]
ax=[2ax-σ(a)x-a]y
y=ax/[2ax-σ(a)x-a]
Then, you put a/[2a-σ(a)]=b/c.
2a-σ(a)=ac/b, σ(a)=2a-(ac/b)
∴
y=ax/[(ac/b)x-a]=bx/(cx-b)
cy-b=c[bx/cx-b]-b=(b^2)/cx-b
b^2=(cx-b)(cy-b)
This square includes prime numbers infinitely.
For example, a=4.
4/[8-σ(4)]=4/(8-7)=4/1=b/c
σ(4)=1+2+4=7
∴
b=4, c=1
16=(x-4)(y-4)
x-4 | y-4 | x | y | p=x-1 | q=y-1 | r=xy-1 |
16 | 1 | 20 | 5 | 19 | 4 (not prime) | |
8 | 2 | 12 | 6 | 11 | 5 | 71 |
Therefore
M=apq=4*11*5=220
N=ar=4*71=284
M and N are amicable, and you see prime numbers in it.
(5020,5564) is also amicable.
5020=2^2*5*251
5564=2^2*13*107
This is prime factorization.
You need to calculate all amicable numbers, but they must exist infinitely because of fractal.
