Are are infinitely many intimate pairs of integers? Using the standard statistical definitions, the variance of $x_1, x_2, \ldots, x_n$ and the squared errors about its mean $\mu$ are given by $\sigma^2 = \sum_i(x_i - \mu)^2/n$ and $\delta_2 = \sum_i(x_i - \mu)^2 = n\sigma^2$ respectively. 

Definition 1: The variance and the squared error of an integer is defined as the variance and the squared error of its positive divisors
  respectively.

I found that: 


*

*There are distinct integer pairs whose variances are equal. The smallest such pair is $(691, 817)$. Let us call them equivarient integers.

*There are integer pairs whose squared errors are equal. The smallest such pair is $(45, 53)$.


The more interesting fact is that there are equivarient pairs which have the same number of divisors and hence their squared errors are also equal. We define:

Definition 2: Two distinct integers are said to be an intimate pair if they are have the same number of divisors and the same
  variance.

The first few intimate pairs are $(1403,1461)$, $(1564,1572)$,$(2068,2076)$,$(2249,2305)$,$(3397,3493)$,$(7871,8193)$,$ (23903,24101)$,$(61769, 64443)$.
Questions:


*

*Are there infinitely many intimate pairs?

*Are there three or more integers that are intimate ( and what should we call them hahaha)


Note: Please let me know in case there is any reference in literature. I could not find any. Posted to MO since it is unanswered in MSE
 A: For what it's worth, not an answer but a long comment:
Let $\sigma_k$ denote the sum of the $k^{th}$ powers of the divisors. 
If we consider an integer with a single prime factor, say
$$p^{n-1},$$
we have
$$\sigma_0=n,$$
$$\sigma_1=\frac{p^n-1}{p-1},$$
$$\sigma_2=\frac{p^{2n}-1}{p^2-1}.$$
Then with the variance $v$,
$$v\,\sigma_0^2=\sigma_0\sigma_2-\sigma_1^2=n\frac{p^{2n}-1}{p^2-1}-\left(\frac{p^n-1}{p-1}\right)^2=(p^n-1)\frac{n(p-1)(p^{n}+1)-(p+1)(p^n-1)}{(p+1)(p-1)^2}.$$
This function seems to be growing for $p>1$ and any $n$, so chances are low that two distinct $p$ yield the same variance.
Now with two prime factors, say
$$p^{n-1}q^{m-1},$$
we have
$$v\,\sigma_0^2=\sigma_0\sigma_2-\sigma_1^2=nm\frac{p^{2n}-1}{p^2-1}\frac{q^{2m}-1}{q^2-1}-\left(\frac{p^n-1}{p-1}\frac{q^m-1}{q-1}\right)^2,$$
which just seems intractable.

In the simplest case, $n=m=2$, the expression reduces to
$$4(p^2+1)(q^2+1)-(p+1)^2(q+1)^2.$$
The challenge is to find two points with prime coordinates on the iso-curves of this function. But from Wolfram Alpha, the only solution is $(1,1)$. I tried a few other exponents, to no avail.
