# 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:

1. Are there infinitely many intimate pairs?
2. 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

• If I am right, you can replace one of "equivarient" or "same error" by "same number of divisors", which is also "same multiplicities in the prime factorization". – Yves Daoust Apr 10 at 7:08
• @YvesDaoust: Valid point, makes it simpler. I am updating it. – Nilotpal Kanti Sinha Apr 10 at 7:11
• Is this a conjecture of yours? (Also, looking at your bio, you say you are a data scientist. To be that requires an extraordinary mind, so a quick well done!) – Mr Pie Apr 10 at 7:20
• @user477343 Does it? Being one too, I'll take that as a compliment. – J.G. Apr 10 at 7:23
• @user477343 Yes, its my conjecture. Thanks for the compliment !!! – Nilotpal Kanti Sinha Apr 10 at 7:26

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.