The equation $\sigma(n)=\sigma(n+1)$

In OEIS, the solutions of $$\sigma(n)=\sigma(n+1)$$ where $\sigma(n)$ denotes the sum of the divisors of $n$ including $1$ and $n$ , are shown upto $n=10^{13}$

The entry can be found already by entering the first three solutions $14,206,957$

It is mentioned that it is unknown whether there are infinite many solutions.

My questions :

• Has $$\sigma(n)=\sigma(n+1)=\sigma(n+2)$$ a solution ? I checked the OEIS-entries and upto $10^{13}$ there is none. Can we perhaps show that this double-equation cannot have a solution ?

• Is a family of solutions known that is probably infinite (but not proven, since the problem whether infinite many solutions exist is open) ?

• Can we find the next solution more efficiently than by just checking all cases ?

Solution 1:

Just sharing some ideas that could be useful for resolving this problem, and which are too long to fit in the Comments section:

• The expression $n(n+1)(n+2)$ is divisible by $6$, which is an even perfect number.
• We can use the divisibility constraint $\gcd(n,n+1)=\gcd(n+1,n+2)=1$, and $\gcd(n,n+2)=1$ (if $n$ is odd).
• We can then apply the $\sigma$-function to the products $$n(n+1)$$ $$(n+1)(n+2)$$ and $n(n+2)$ (if $n$ is odd in this last case).
• Then note that we have $$\sigma(n(n+1))=\sigma(n)\sigma(n+1)=\sigma((n+1)(n+2))=\sigma(n+1)\sigma(n+2)=k^2$$ where $k=\sigma(n)=\sigma(n+1)=\sigma(n+2)$ is the common value.
• If $n$ is odd, note that we have $$\sigma(n(n+2))=\sigma(n)\sigma(n+2)=k^2.$$
• So now, consider the entire product $n(n+1)(n+2)$. This is a nontrivial multiple of $6$ if $n > 1$, so that $$\frac{\sigma(n(n+1)(n+2))}{n(n+1)(n+2)}>2.$$ Assume to the contrary that $n>1$ is odd. Then the last inequality implies $$\frac{\sigma(n)}{n}\frac{\sigma(n+1)}{n+1}\frac{\sigma(n+2)}{n+2}>2,$$ so that we obtain $$k^3=\sigma(n)\sigma(n+1)\sigma(n+2)>2n(n+1)(n+2).$$ We therefore get the inequality $$\sigma(n)=\sigma(n+1)=\sigma(n+2)>\sqrt[3]{2n(n+1)(n+2)}.$$

I will stop here.

Solution 2:

Comment:

Here I consider a particular case .

$N=p_1p_2$ ⇒ $\sigma_N=p_1+p_2+p_1p_2+1$

Where $P_1$ and $p_2$ are primes.

Suppose $p_1p_2 +p_1+p_2=p$ is prime, then:

$\sigma_p=\sigma_N=p_1+p_2+p_1p_2+1=p+1 $

There can be infinitely many couples of this type, for example:

$p_1=2$, $p_2=13$ gives $N=26$ and $p=41$, or $p_1=3$ and $p_2=17$ gives $N=51$ and $p=71$

Now consider couple (33, 35), we try to find a third number with sum of divisors equal to that of $33$:

$N=33=3\times 11$ ⇒ $\sigma_N=3+11+33+1=48$

$p=3+11+33=47$ ⇒ $\sigma_p=47+1=48$

$\sigma_{q=35}=5+7+35+1=48$

so we have a triple $(33, 35, 47)$

There can be infinitely many triples of this type too.This experiment shows that the probability of existence of a triple with small difference is not zero if N is large enough.

For $\sigma(n)=\sigma(n+1)=\sigma(n+2)$, if we want to use this model, we have to look for a large number like $N=p_1p_2p_3 . . .$such that it gives numerous options for combination to construct two new numbers,the difference between three numbers $N$, $p$ and $q$ is as small as possible.This difference might never be equal especially equal to $1$. algorithm can be more efficient if one of these numbers is prime.

UPDATE:

I checked this by a simple proram which finds integer solutions for x and y in following Diophantine equation:

$xy+x+y=p$

Where p is a primes which is the known of equation. This is the results:

1- Some primes such as $9973, 9967, 9949, 9817, 9721 . . .$ give no solution.

2- Some primes such as $9923$ give only one prime solution for x and y:

$x=2, y=3307$ such that: $9923=2\times 3307+2+3307$ and we have:

$\sigma_{9923}=\sigma_{2\times 3307=6614}=9924$

3-Some primes give more than one prime solution for x and y:

$71$ gives: $(x, y, N)=(2, 23, 46), (3, 17, 51), (5, 11, 55)$

The differences are interesting:

$55-51=4$

$51-46=5$

$9719$ gives: $(x, y, N)=(89, 107, 9523),(53, 179, 9487), (11, 809, 8899), (5, 1619, 8095)$

The program is simple for Python:

for x in range (2, 5000):

 for y in range (2, 5000):

      if x*y+x+y==p:

          print x, y

I gave p manually. I am not good enough to write a program that my computer do it automatically, also my computer is too weak for this calculations. But I am quite sure we can find triples or quadruple or even family of numbers with larger number of members which have equal sum of divisors, among these members some make an arithmetic progression.