13
$\begingroup$

Consider the prime factorization of the numbers $14$ and $15$ :

$$14 = 2 \cdot 7 \implies \tau(14) = 2 \cdot 2 = 4 \space ;\space \sigma(14) = 3 \cdot 8 = 24$$ $$15=3 \cdot 5 \implies \tau(15) = 2 \cdot 2 = 4 \space ;\space \sigma(15) = 4 \cdot 6 = 24$$ Thus, we can note that $\tau(14) = \tau(15)$ and $\sigma(14) = \sigma(15)$ . Here, $\tau(x)$ is the number of positive divisors of $x$ and $\sigma(x)$ is the sum of positive divisors of $x$.

There are many examples such as $\{14,15\}$ that show the above properties. Another example is $\{33,35\}$ and $\{46,51,55\}$. More than just pairs or triples, we can have even larger sets. The smallest sextuplet (set of six positive integers) is $\{282,310,322,345,357,385\}$.

We can see that as $\{14,15\}$ have their union of prime factors as $\{2,3,5,7\}$ and since the $\tau$ and $\sigma$ functions are multiplicative functions, we can take any positive integer $k$ such that $\gcd(k,210) = 1$ and then the family of solutions $\{14k,15k\}$ would satisfy the properties. Thus, there are infinitely many pairs of distinct positive integers with equal number of divisors and equal sum of divisors.

However, the question gets more interesting if we ask whether there are infinitely many such pairs with relatively prime positive integers, i.e. whether there are infinitely many such families of solutions.

The 'relatively prime' part is removed to untrivialize the problem. We could however look at it at a different angle. The problem is only trivial if we could multiply equal powers of primes $p$. Thus, we could also ask if there are infinitely many pairs of positive integers $\{a,b\}$ such that $\tau(a) = \tau(b)$ and $\sigma(a) = \sigma(b)$, with the condition that for no prime $p$, $\nu_p(a) = \nu_p(b) \neq 0$.

This problem is deeply related with conjectures such as Dickson's conjecture and Schinzel's Hypothesis. However, it is much weaker than such problems. Yet, when we concentrate on certain cases, for example, $\tau(a)=\tau(b)=4$ where $a,b$ are both products of two distinct primes, we get for primes $p,q,r,s$ : $$(p+1)(q+1) = (r+1)(s+1)$$ By fixing $s=7$ and $q=3$ for example , we get $p+1 = 2r+2$ which gives that for primes $p,r$ ; we have $p=2r+1$. Thus, we are taken to the Sophie-Germain problem. Our problem is solved directly by solving any such single cases of Dickson's conjecture.

Primes approximately have $0.5$ probability to be $1 \pmod{4}$ and $0.5$ probability to be $3 \pmod{4}$. Thus, it is expected that $8 \mid (p+1)(q+1)$. Similarly, primes approximately have $0.5$ probability to be $1 \pmod{3}$ and $0.5$ probability to be $2 \pmod{3}$. Thus, it is expected that $3 \mid (p+1)(q+1)$. Hence, it is better to search for $(p+1)(q+1) = (r+1)(s+1) = 24k$.

It can be seen quite clearly that even such small cases of the problem are likely to generate infinitely many solutions, like the above. Thus, it seems most likely that the answer to this question is affirmative.

We can extend our problem to not only pairs, but to sets of distinct positive integers of size $n$. We can also extend our problem by asking whether there are arbitrarily large sets of distinct positive integers with equal number of divisors and equal sum of divisors. This seems likely from the example of the sextuplet. This also directly follows from the full-fledged version of Dickson's conjecture. Finally, this is a stronger version of our previous question as we cannot have infinitely many finite positive integers in a set as they would have finite value for $\sigma(x) \geqslant x$, which would bound the size of the set.

Any suggestions and progress would be accepted. Please share any ideas/remarks in the comments. Ideas can be shared on any of the problems above.

$\endgroup$
5
$\begingroup$

A quick computational note on (what I see as) the main question raised:

Do there exist infinitely many pairs of natural numbers $\{a,b\}$ such that $\tau(a)=\tau(b)$ and $\sigma(a)=\sigma(b)$ with the property that $\forall p\left( \nu_p(a)=\nu_p(b)\Rightarrow\nu_p(a)=\nu_p(b)=0\right)$?

I will be truly shocked if this answer is negative; simply by guessing and checking I have found quadruples of distinct primes $(p,q,r,s)$ such that $$ (p+1)(q+1)=24k=(r+1)(s+1) $$ for no less than $12$ values of $k$. I have listed them below:

$k=1\quad(3,5,2,7)\\k=2\quad(5,7,3,11)\\k=3\quad(5,11,3,17)\\k=4\quad(7,11,2,31)\\k=5\quad(5,19,3,29)\\k=6\quad(7,17,2,47)\\k=7\quad(11,13,3,41)\\k=8\quad(7,23,5,31)\\k=9\quad(11,17,3,53)\\k=10\quad(11,19,7,29)\\k=12\quad(11,23,5,47)\\k=14\quad(13,23,7,41)$

It seems to me that you are figuratively tripping over solutions to this problem. I would suggest trying to approach this question not through specifications into known conjectures like Dickson's Conjecture or Schinzel's Hypothesis, but rather by seeing how developed the literature is in solving Diophantine equations in the primes, like this paper does.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.