0

Question: If $p^n$ divides product of two consecutive positive integers say, $m(m-1)$ then $\;$ $p^n\lvert\;{m}$ $\;$or $\;$ $p^n\;\lvert\;{m-1}$.

For example

$2^3\lvert\;m(m-1)$ then $2^3\;\lvert\;m$ $\;$ or $\;$ $2^3\;\lvert\;m-1$

My attempt: By properties of primes, I know that, if the prime number $p$ divides product of two primes say $ab$ then $p$ divides $a$ or $p$ divides $b$.

But here in our case, we have $p^n$ & $n$ may be greater than $1$. So I think, this may not be true. But on other hand, I could not find any counterexample.

Please help.

Bill Dubuque
  • 282,220
General Mathematics
  • 376
  • 4
    If $p^n\mid (m-1)m$, then $p\mid (m-1)m$, so either $p\mid m$ or $p\mid (m-1)$, but not both. So assume $p\mid m$. Then also $p^n\mid m$ since $p\nmid m-1$. – Dietrich Burde Jul 23 '24 at 14:19
  • @DietrichBurde sir, $5$ divides $35$ but, $5^2$ does not divides $35$ (even though $5$ does not divides $34$) – General Mathematics Jul 23 '24 at 14:26
  • 2
    But in your case $5^2$ doesn't divide $34\cdot 35$, as you have assumed. You said, "if $p^n$ divides $m(m-1)$". – Dietrich Burde Jul 23 '24 at 14:27
  • Yes. I forgot. Sir, but how can we prove it in standard way? – General Mathematics Jul 23 '24 at 14:29
  • 2
    This is the standard way. We only use that $p\mid ab$ implies $p\mid a$ or $p\mid b$ and that $p\mid a$ and $p\mid a-1$ implies $p\mid 1$, a contradiction. These are standard arguments of elementary number theory. – Dietrich Burde Jul 23 '24 at 14:30
  • 1
    A more general statement is true (and perhaps easier to see how to prove): If $p^n$ divides $ab$ and $\gcd(a,b)=1$, then $p^n\mid a$ or $p^m\mid b$. – Greg Martin Jul 23 '24 at 16:43
  • By the dupe, if $,(m,m')!=!1,$ and $,p^n\mid mm',$ then $,p^j\mid m,,p^k\mid m',,$ with $, j+k=n,,$ so $,(m,m')!=!1\Rightarrow j=0,$ or $,k=0,,$ so $,k=n,,$ or $,j=n.\ \ $ – Bill Dubuque Jul 23 '24 at 22:36
  • Or prime $,p\mid p^n\mid mm'\Rightarrow p\mid m,$ or $,m';,$ wlog $,p\mid m,,$ so $,(m,m')!=!1\Rightarrow (p,m')!=!1\Rightarrow \color{#c00}{(p^n,m')!=!1},,$ so $,\color{#c00}{p^n\mid m'}m\Rightarrow p^n\mid m,$ by Euclid's Lemma .$\ \ $ – Bill Dubuque Jul 23 '24 at 23:01
  • Key Idea by the first dupe any divisor $,\rm d,$ of a product of coprimes $mm'$ must itself split into a product of coprimes $, {\rm d} = dd',$ where $,d\mid m,,d'\mid m'.,$ But when $,{\rm d}= p^n,$ its only coprime splittings are when one factor is $1$ and the other is $,\rm d.\ \ $ – Bill Dubuque Jul 23 '24 at 23:16

0 Answers0