Amicable numbers

The smallest pair of amicable numbers is (220, 284). You expand it infinitely.

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.

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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.

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You need to calculate all amicable numbers, but they must exist infinitely because of fractal.