# $1$ as difference of composites with same number of prime factors and smallest examples

It is probably open can we for every $k \in \mathbb N$ find two composites $a_k$ and $b_k$ such that $a_k$ and $b_k$ have exactly $k$ prime factors and $a_k-b_k=1$.

Smallest examples found so far are:

for $k=1$ $$3^2-2^3=1$$ for $k=2$ $$3 \cdot 5 - 7 \cdot 2=1$$ for $k=3$ $$3\cdot7\cdot11-2\cdot5\cdot23=1$$ for $k=4$ $$5 \cdot 7 \cdot 11 \cdot 19 - 2 \cdot 3 \cdot 23 \cdot 53=1$$ for $k=5$ $$3\cdot7\cdot17\cdot23\cdot31-2^2\cdot5\cdot11\cdot13\cdot89=1$$

It is easy to see that $3$ is a factor of one of numbers that are smallest pair for $k=1,2,3,4,5$ so I will make a pretty dumb conjecture that could be ruled out with a clever computer-check:

If a pair $(a_k,b_k)$ is smallest pair then $3$ is a factor of one of those two numbers.

Until which $k$ is this true?

I am prepared to let this always be true because smallest pairs should tend to have small primes as factors of them and because there is a good chance that one of members is divisible by $3$ because they differ only by one.

I am not sure would I like to see a counterexample, but go for it.

This is true at least for $k=1,2,3,4,5,6,7,8,9,10,11$ by this list .

A same question asked on MSE where there is an answer by Oleg567 proposing smallest examples for $k=12,13,14,15,16$

You start with "It is probably open can we for every $k\in \mathbb{N}$ find two composites $a_k$ and $b_k$ such that $a_k$ and $b_k$ have exactly $k$ prime factors and $a_k−b_k=1$."

I will comment on this, but I do not comment on your question in the grey box, about the prime factor $3$.

For $k=1$, observe that there is only one solution of $a^x-b^y=1$, namely $3^2-2^3=1$. This is Catalan's conjecture, proved by Mihailescu.

For $k\geq 3$, there is nice paper "Small gaps between almost primes, the parity problem and some conjectures of Erdos on consecutive integers" by D. A. GOLDSTON, S. W. GRAHAM, J. PINTZ AND C. Y. YILDIRIM. International Mathematics Research Notices, Vol. 2011, No. 7, pp. 1439–1450 http://www.math.boun.edu.tr/instructors/yildirim/paper/GGPY3imrn.pdf

with many interesting results:

Theorem 1.1. Let $A$ be any multiset of positive integers that contains $\{2, 1, 1, 1\}$ as a subset. There exist infinitely many integers $x$ such that $x$ and $x + 1$ both have exponent pattern $A$. Consequently, for any integer $B \geq 0$, there exist infinitely many integers $x$ such that $\omega(x) = \omega(x + 1) = 4 + B, \Omega(x) = \Omega(x + 1) = 5 + B$, and $d(x) = d(x + 1) = 24 \, 2^B$. ...

Theorem 1.2. There are infinitely many integers $x$ with $\omega(x) = \omega(x + 1) = 3$.

Theorem 1.3. There are infinitely many integers x with $\Omega(x) = \Omega(x + 1) = 4$.

And many similar results.

From this it seems that only the case $k=2$ of your question is left. You give one solution, which settles your question.

The more difficult question if there are infinitely many solutions with $k=2$ may be open. But let me mention a partial result:

D. A. Goldston, S. W. Graham, J. Pintz and C. Y. Yildirim Small gaps between primes or almost primes Journal: Trans. Amer. Math. Soc. 361 (2009), 5285-5330 https://www.ams.org/journals/tran/2009-361-10/S0002-9947-09-04788-6/home.html

Theorem 3. Let $q_n$ denote the $n$-th number that is a product of exactly two primes. Then $\lim \inf_{n \rightarrow \infty} (q_{n+1} - q_n) \leq 26$.