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I noticed and found only first three cases:

We can write $1$ as difference of two composites that have one prime factor $$3^2-2^3=1$$

and as difference of two composites that have two prime factors $$3\cdot 5 - 7\cdot 2 = 1$$

and as difference of two composites that have three prime factors $$2^2 \cdot 3^2 \cdot 43-7 \cdot 13 \cdot 17=1$$

I believe that this holds for every $k \in \mathbb N$, that is, that for every $k \in \mathbb N$ there exist composites $a_k$ and $b_k$ that have exactly $k$ prime factors and are such that we have $a_k-b_k=1$.

Is my belief true? Is this known? What is known about all of this and similar problems? Can someone find solutions for some larger $k$´s?

Examples exist at least for $k=1,2,...11$ by this list .

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  • $\begingroup$ Also asked on MSE: math.stackexchange.com/questions/2700917/… $\endgroup$
    – Shalom
    Commented Mar 22, 2018 at 8:58
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    $\begingroup$ Why close? This seems to be pretty research level. $\endgroup$
    – Fedor Petrov
    Commented Mar 22, 2018 at 9:49
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    $\begingroup$ The smallest example for $k=3$ is $3\cdot7\cdot11-2\cdot5\cdot23=231-230=1$. $\endgroup$
    – Gerry Myerson
    Commented Mar 22, 2018 at 12:03
  • $\begingroup$ @GerryMyerson And the smallest examples, so far, for $k=1,2,3$, all three have $3$ as a factor of a bigger number, and $2$ as a factor of a smaller number. $\endgroup$
    – Shalom
    Commented Mar 22, 2018 at 12:08
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    $\begingroup$ The smallest example for $k=4$ is $7315-7314=1$ (and 3 is not a factor of 7315). $\endgroup$
    – Gerry Myerson
    Commented Mar 22, 2018 at 12:11

1 Answer 1

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Erdos mentions in his book "Topics in the Theory of Numbers" the following:

"It is stil unkown if the equation $\omega(n+1)=\omega(n)$ has infinitely many solutions...It is known that $|\omega(n+1)-\omega(n)|$ takes a certain value infinitely often".

Here $\omega(x)$ denotes the number of prime factors of $x$.

So, I suppose this is still open. I am not sure if there is an update on this problem.

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  • $\begingroup$ Do you know is $\omega(n)=\omega(n+k)$ open for every $k \in \mathbb N$, in the sense of infinity of solutions? $\endgroup$
    – Shalom
    Commented Mar 22, 2018 at 10:19
  • $\begingroup$ @Shalom Unfortunately no. $\endgroup$
    – Konstantinos Gaitanas
    Commented Mar 22, 2018 at 20:05
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    $\begingroup$ Actually, it is by now known. see: arxiv.org/pdf/1105.1621v1.pdf $\endgroup$
    – Pablo
    Commented Mar 23, 2018 at 6:47

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