What is $\{a,b\}$?

The elements $a$ and $b$ ?o_0

SPHS1234 wrote: The elements $a$ and $b$ ?o_0 ty

Chestnut Ramification Theorem go brrrrr. Without loss of generality let $p<q$ and let $2<p_1<p_2<\cdots<p_k<p<p_{k+1}<p_{k+2}<\cdots<p_\ell<q$ be the distinct odd primes, not $p$ or $q$, strictly less than $q$. Then, we can choose by the Chestnut Ramification Theorem \[m\equiv 1\pmod 2\]\[m\equiv 0\pmod p_j\qquad 1\leq j\leq \ell\]\[m\equiv -2\pmod p\]\[m\equiv -4\pmod q\]Now, if we take $n=m+2$, we can see it is odd, not divisible by $p_j$ for $1\leq j\leq\ell$, and is divisible by $p$. Thus, $d(n)=p$. By similar logic, $d(n+2)=q$.

Can we say something similar about largest prime divisors?

llplp wrote: Can we say something similar about largest prime divisors? That was originally the intent, but we did not know how to deal with such statement with no restrictions on $p,q$.

Not too related but here's a similar result with some restrictions on $p,q$, and consecutive instead of $n,n + 2$. Tuymaada 2016 Senior, P5 wrote: The ratio of prime numbers $p$ and $q$ does not exceed 2 ($p\ne q$). Prove that there are two consecutive positive integers such that the largest prime divisor of one of them is $p$ and that of the other is $q$.

GorgonMathDota wrote: Not too related but here's a similar result with some restrictions on $p,q$, and consecutive instead of $n,n + 2$. Tuymaada 2016 Senior, P5 wrote: The ratio of prime numbers $p$ and $q$ does not exceed 2 ($p\ne q$). Prove that there are two consecutive positive integers such that the largest prime divisor of one of them is $p$ and that of the other is $q$. Yep that was the original idea of ours too, but we soon realized that this appeared on AoPS for so many times already lol.

I am quite confused on this problem. This is my try for a construction. Let $n=p \times Q$ where $p$ is the smallest prime divisor of $n$ and $Q \in \mathbb{N}$. Clearly, $d(n)=p$. Also, $n+2=pQ+2$. We want $d(pQ+2)=q$. For this we wish to show that such $Q$ exists. It is well known that the solution to $pQ+2 \equiv 0 \pmod{q}$ exists. As we want that $q$ should be the smallest prime dividing $pQ+2$, we define a sequence of primes $x_1,x_2 \ldots x_m$ such that: \[2<x_1<x_2 \ldots < x_m<q\]and $x_i \mid pQ+2$ where $1 \le i \le m$. Now we don't want the following set of congruences to hold true: \begin{align*} pQ+2 &\equiv 0 \pmod{x_1} \\ pQ+2 &\equiv 0 \pmod{x_2} \\ &\vdots \\ pQ+2 &\equiv 0 \pmod{x_m} \end{align*}From here on, I think it is possible to pick such $Q$ which doesn't satisfy any of the above congruences and is a solution to $pQ+2 \equiv 0 \pmod{q}$. As soon as we pick such $Q$, we are done. $\blacksquare$ Is this argument fine? Or it should be more rigorous? Is picking such $Q$ is so easy? Basically, is my solution correct? I personally think there should be more details, but I am not able to figure it out. Can I get some help?

Just stating that $Q$ exists is wrong,otherwise what is left in the problem Let's find $Q$ Assume $q>p$ Let the $x_{i}s$ exclude $p$. Then solve the following congruences. $pQ \equiv 2 \pmod{x_{1}....x_{m}}$ and $pQ \equiv -2 \pmod{q}$ After the Chinese Remainder Theorem, we multiply with the inverse of $p$ $\pmod {q.x_{1}....x_{m}}$ to get $Q$. Now, note that since none of the primes is $2$, $2$ is never congruent to $-2$. Also, $rad(Q)$ doesnt contain a prime smaller than $p$ as none of the $x_{i}s$ divide it.Also choose $Q$ odd. (By the way, how is ur Himalaya meditation going on).

naman12 wrote: Chestnut Ramification Theorem go brrrrr. Without loss of generality let $p<q$ and let $2<p_1<p_2<\cdots<p_k<p<p_{k+1}<p_{k+2}<\cdots<p_\ell<q$ be the distinct odd primes, not $p$ or $q$, strictly less than $q$. Then, we can choose by the Chestnut Ramification Theorem \[m\equiv 1\pmod 2\]\[m\equiv 0\pmod p_j\qquad 1\leq j\leq \ell\]\[m\equiv -2\pmod p\]\[m\equiv -4\pmod q\]Now, if we take $n=m+2$, we can see it is odd, not divisible by $p_j$ for $1\leq j\leq\ell$, and is divisible by $p$. Thus, $d(n)=p$. By similar logic, $d(n+2)=q$. Can you link a statement of the theorem you have made use of? I tried googling it but can't seem to find anything close to what you have used.

starchan wrote: naman12 wrote: Chestnut Ramification Theorem go brrrrr. Without loss of generality let $p<q$ and let $2<p_1<p_2<\cdots<p_k<p<p_{k+1}<p_{k+2}<\cdots<p_\ell<q$ be the distinct odd primes, not $p$ or $q$, strictly less than $q$. Then, we can choose by the Chestnut Ramification Theorem \[m\equiv 1\pmod 2\]\[m\equiv 0\pmod p_j\qquad 1\leq j\leq \ell\]\[m\equiv -2\pmod p\]\[m\equiv -4\pmod q\]Now, if we take $n=m+2$, we can see it is odd, not divisible by $p_j$ for $1\leq j\leq\ell$, and is divisible by $p$. Thus, $d(n)=p$. By similar logic, $d(n+2)=q$. Can you link a statement of the theorem you have made use of? I tried googling it but can't seem to find anything close to what you have used. I think Chestnut Ramification Theorem and Chinese Remainder Theorem are same , because I didn't anything other than CRT in the solution..

starchan wrote: naman12 wrote: Chestnut Ramification Theorem go brrrrr. Without loss of generality let $p<q$ and let $2<p_1<p_2<\cdots<p_k<p<p_{k+1}<p_{k+2}<\cdots<p_\ell<q$ be the distinct odd primes, not $p$ or $q$, strictly less than $q$. Then, we can choose by the Chestnut Ramification Theorem \[m\equiv 1\pmod 2\]\[m\equiv 0\pmod p_j\qquad 1\leq j\leq \ell\]\[m\equiv -2\pmod p\]\[m\equiv -4\pmod q\]Now, if we take $n=m+2$, we can see it is odd, not divisible by $p_j$ for $1\leq j\leq\ell$, and is divisible by $p$. Thus, $d(n)=p$. By similar logic, $d(n+2)=q$. Can you link a statement of the theorem you have made use of? I tried googling it but can't seem to find anything close to what you have used. I guess he was just trying to make a joke with CRT

OK I officially got trolled in the worst way possible. Still it's surprising my brain did not realise it was basically just CRT.

Wait this is just CRT isn't it? Let $$S=\{p_1,p_2, \dots, \max{(p, q)}\}-\{p, q\}$$be the set of primes less than and distinct to $p, q$. Take $n$ such that $n\equiv 0 \pmod p$, $n \equiv -2 \pmod q$ and $n \not\equiv 0, -2 \pmod r$ for every $r\in S$. This has a solution by CRT so we are done, since this implies $d(n)=p$ and $d(n+2)=q$

WLOG let $p > q$. By CRT, there exists a positive integer $n$ such that $n\equiv 0\pmod{p}$ $n\equiv -2\pmod{q}$ and $n\equiv -1\pmod{r}$ for all primes $r < p$, $r\neq q$. This $n$ clearly satisfies the condition in the problem statement.

CRT and Drichlet theorem. $p\ge q$. $r$ is prime number. $$n=(p^x)r+2$$$A$ - all prime numbers products less then $q$. $$x=\phi(Aq)$$$$r\equiv -2\pmod{q}$$$$r\equiv -1\pmod{A}$$exist $r$ prime great $p$.

By CRT choose $n \equiv 0 \pmod p, -2 \pmod q$, $1$ modulo all other primes less than $\max(p,q)$ except $3$, and $2 \pmod 3$. This clearly works.