I recently saw a video stating that if the Collatz conjecture was proven for prime numbers, it was proven for all numbers. That's the first time I see this and I can't find any references with the classical JFGI.
It says something like this ($a$, $b$, $n$ are odd integers):
$\sigma_p(n)$ is the Collatz trajectory ($p$ steps applied to $n$) and for a fixed function $f(b)$ not provided you have:
$\sigma_p(an+b)=a\cdot\sigma_p(n)+f^p(b)$
and specially for odd prime factorization:
$\sigma_p(an)=a\cdot\sigma_p(n)+f^p(0)$
Apparently, it can be shown that if $\sigma(n)<n$ than $\sigma(an)<an$ and it suffice to show for all prime $n$ since a finite stopping time for primes implies a finite stopping time for all its composite
I tried on some numbers (e.g., $85=5\cdot 17$) but couldn't figure it out.
Is there a paper on this subject? The video seems to talk about a $15$ page proof.
NEW SECTION ON THE BOTTOM
I think it is highly likely that there is no paper showing that if the collatz conjecture is true for primes it is true for all numbers.
In the original post, it was claimed that it can be shown that $\sigma\left( n\right) \lt n \implies \sigma\left( an\right) \lt an$
( I'm assuming the implication is $\sigma_p\left( n\right) \lt n \implies \sigma_p\left( an\right) \lt an \quad$ and not $\sigma_p\left( n\right) \lt n \implies \sigma_q\left( an\right) \lt an$ )
This implication is not only false for specific cases, it is also false for whole families of cases.
If $n$ is of the form $4b+1$ and $a$ is of the form $4c+3$ then $\sigma_3\left( n\right) \lt n$ but $\sigma_3\left( an\right) \gt an$
On top of that, the factorization of a number doesn't help that much in determining the trajectory of the number in collatz conjecture. There are three operators that are used in the conjecture (/2,*3,+1). The /2 and the *3 operators change the factorization of the number by adding or subtracting a single prime factor. The +1 operator however changes the factorization of the number in unpredictable ways. There are two things that you can conclude after the +1. The first is that since the number was odd before the +1 it will be even after the +1. The second is that since two consecutive numbers don't share any factors, the number before +1 will have a different set of factors from the number after +1. After those two things any information about the factorization of the numbers, in general, is really hard to come by. This is what makes collatz conjecture so difficult, to begin with. The idea that you can gain information about the trajectory of multiples of primes by learning about the trajectory of prime values just doesn't make sense.
$31$ takes $106$ steps to reach $1$, but $31*11$ takes $11$ steps to reach $1$
$3$ takes $7$ steps to reach $1$, but $3*9$ takes $111$ steps to reach $1$
$17$ takes $12$ steps to reach $1$, but $17*40$ takes $12$ steps to reach $1$
So in the first case multiplying a prime by a number decreased the number of steps by a large margin. In the second case multiplying a prime by a number increased the number of steps by a large margin. In the third case multiplying a prime by a number kept the number of steps the same.
Edit: The OP has said in comments that the person in the video said that the function is an affine function. (Which I probably should have figured out beforehand). This means that $f$ is of the form $ f(x) = mx+n$ where $m,n\in \Bbb{R}$. $m,n\in \Bbb{R}$ for general affine equations. In this case, $m,n\in \Bbb{Q}$ because Rational numbers are closed under the four basic operators ($+,-,*,/$).
The OP mentioned in the original post that one of the equations used in the video is:
$\sigma_p(an)=a\cdot\sigma_p(n)+f^p(0)$
This means that $f^p(0)$ is just a constant. One thing I realized is that $\sigma_p(an)$ and $\sigma_p(n)$ are discordant. Not only because one is by itself and one is multiplied by $a$ but they don't perform the same operations from step to step. One may do the odd operation $(*3+1)$ while the other may do the even step $(/2)$. This means that $f^p(0)$ is just the correction term needed to make the equation equal. This is why $f$ isn't defined. It is almost impossible to know what $f^p(0)$ will be without calculating $\sigma_p(an)$ and $\sigma_p(n)$ then solving for $f^p(0)$. We can see how this would work by using one of the examples above. Let $n=31$ and $a=11$.
We start with $\sigma_0(11*31)=11\cdot\sigma_0(31)+f^0(0)$. This means we haven't applied any steps so $f^0(0)=0$ and the equation is just $341=341$. Next,
$\sigma_1(11*31)=11\cdot\sigma_1(31)+f^1(0)$
$\sigma_1(11*31)=1024$, $\quad\sigma_1(31)=94$
both sigmas apply an odd step. Multiplying by three does the same thing to both sides but, adding $1$ to both sides does not. On the left side, one is added but the one on the right is multiplied by 11 so this adds $11$. This means $f^1(0)=-10$ Next,
$\sigma_2(11*31)=11\cdot\sigma_2(31)+f^2(0)$
Both sigmas apply an even step so both sides are divided by $2$ including the $f^1(0)$. So $\sigma_2(11*31)=512,\quad\sigma_2(31)=47,\quad f^2(0)=-5$. Next,
$\sigma_3(11*31)=11\cdot\sigma_3(31)+f^3(0)$
The left sigma applies an even step and the right sigma applies an odd step so $\sigma_3(11*31)=256,\quad\sigma_3(31)=142,\quad \bf{f^3(0)=-1306}$. Next,
$\sigma_4(11*31)=11\cdot\sigma_4(31)+f^4(0)$
$\sigma_4(11*31)=128,\quad\sigma_4(31)=71,\quad f^4(0)=-653$
$\sigma_5(11*31)=11\cdot\sigma_5(31)+f^5(0)$
$\sigma_5(11*31)=64,\quad\sigma_5(31)=214,\quad f^5(0)=-2290$
If $f$ is just a post-hoc correction term it is useless. If $f$ isn't just a post-hoc correction it needs a far more extensive explanation by the person in the video.
EDIT 2:
Earlier in this post, I showed that $\sigma_p\left( n\right) \lt n \nRightarrow \sigma_p\left( an\right) \lt an$. What I want to show here is for non-trival pairs of $p$ and $q\quad \sigma_p\left( n\right) \lt n \nRightarrow \sigma_q\left( an\right) \lt an$.
The trivial case of the implication $\sigma_1\left( n\right) \lt n \implies \sigma_1\left( an\right) \lt an$ is true. $\sigma_1\left( n\right) \lt n$ implies that $n$ is even. Any integer multiplied by an even number is even therefore $an$ is also even. So after this paragraph $a$ and $n$ are odd.
In order to show that the implication is false for non-trivial $p$,$q$ pairs I will use Euler’s theorem and an expansion property of a specific type of number ($x^{yz}-1$) where $x,y,z\in\Bbb{N}\space|\space x,y,z>1$. Euler’s theorem states that if $r,s \in \Bbb{N}\space|\space gcd(r,s)=1$ then $s\space |\space r^{\phi\left(s\right)}-1$ Where $\phi()$ is the Euler totient function. The expansion property is $x^{yz}-1=(x^y-1)(x^{y(z-1)}+x^{y(z-2)}+x^{y(z-3)}+…+x^{3y}+x^{2y}+x^y+1)$ this implies $(x^y-1)|(x^{yz}-1)$
After $2\phi(n)$ collatz steps the expression $2^{\phi(n)}-1$ turns into $3^{\phi(n)}-1$ without going lower than $2^{\phi(n)}-1$. This means that $\sigma_{2\phi(n)}\left( 2^{\phi(n)}-1\right) \gt 2^{\phi(n)}-1$. From Euler’s theorem $n|2^{\phi(n)}-1$. Let $k\in\Bbb{N}$. From the expansion property $(2^{\phi(n)}-1)|(2^{k\phi(n)}-1)$. Let $an=(2^{k\phi(n)}-1)$. The expression $2^{k\phi(n)}-1$ is also divisible by $n$ because it contains all factors of $(2^{\phi(n)}-1)$. The expression $2^{k\phi(n)}-1$ takes $2k\phi(n)$ collatz steps to turn into $3^{k\phi(n)}-1$ without going lower than $2^{k\phi(n)}-1$. $k$ can be arbitrarily large this means that with a given $n$ and $q\space \exists\space a,k\space|\space \sigma_q\left( an\right) \gt an$.
For example let’s pick $n=77$ and $q=1000$. $77$ takes three steps to become less than itself. So $\sigma_3\left( 77\right) \lt 77$. In order to make the implication false we need $2k\phi(n)>q\rightarrow k\ge 9$. We let $an=2^{k\phi(n)}-1=2^{540}-1$. This means $\sigma_3\left( 77\right) \lt 77 \nRightarrow \sigma_{1000}\left( 2^{540}-1\right) \lt 2^{540}-1$
$\mathbf{\sigma_p\left( n\right) \lt n \nRightarrow \sigma_q\left( an\right) \lt an}$