Are there pairs of consecutive integers with the same sum of factors? Background/Motivation
I was planning to explain Ruth-Aaron pairs to my son, but it took me a few moments to remember the definition.  Along the way, I thought of the mis-definition, a pair of consecutive numbers with the same sum of divisors.  Well, that's actually two definitions, depending on whether you are looking only at proper divisors.  Suppose all divisors.  I quickly found (14,15) which both have a divisor sum (sigma function) of 24.  Some more work provided (206,207) and then a search on OEIS gave sequence A002961.  
What about proper divisors only?  (2,3) comes quickly, but then nothing for a while.  Noting that the parity of this value ($\sigma(n) -n$) is the same as that of $n$ unless $n$ is a square or twice a square, any solution pair must include one number of that form.  With that much information in hand, I posted this problem at the reference desk on Wikipedia.  User PrimeHunter determined that there were no solutions up to $10^{12}$, but there were no general responses.  
Aside from the parity issue, I haven't found other individual constraints that would filter the candidates--the number of adjacent values identical modulo $p$ for other small primes is at least as great as would be expected by chance, and there are a fair number of pairs that are arithmetically close.  

Other than (2,3), are there pairs of consecutive integers such that $\sigma(n)-n = \sigma(n+1)-(n+1)$?

 A: You should look at Carl Pomerance's follow-up paper: Ruth-Aaron pairs revisited, http://www.math.dartmouth.edu/~carlp/PDF/paper130.pdf .  In his first paper with Erdös they proved a result which showed that the number of RA pairs had asymptotic density 0, but just barely.  In the follow-up Pomerance shows that the the sum of the reciprocals converges (which is much stronger).
A: The question can be rephrased as asking for sigma(n + 1) = sigma(n) + sigma(1), in line with the "Freshman's Dream." 
A: Apologies that this isn't a complete answer. The condition $\sigma(n+1)=\sigma(n)+1$ means that $\sigma(n),\sigma(n+1)$ are relatively prime. In the question, you've taken care of the divisibility by $2$ part. 
Let $p$ be an odd prime and $q>p$ prime. Let $a_{q,p}=a_q=p$ if $q\equiv 1 \bmod p$ and $a_{q,p}=a_q$ equals the order of $q$ modulo $p$ otherwise. If $\nu_q(n)$, the highest power of $q$ dividing $n$ is, $-1$ mod $a_q$, then $p$ divides $\sigma(n)$. So it is necessary that if $q_1,q_2$ are some primes, then one musn't simultaneously have $\nu_{q_1}(n)\equiv -1 \bmod a_{q_1,p}$ and $\nu_{q_2}(n+1)\equiv -1 \bmod a_{q_2,p}$ for any $p,q_1,q_2$.
For example, if $n=q^2$, where $q$ is prime and $1$ mod $3$, then $n+1$ is $2\bmod 4$ and $3$ divides both $\sigma(n), \sigma(n+1)$, so this is an example of $n$ that must be excluded.
