1

Could you give a hint how to prove this?

if $n>5$ is prime, prove $(n-1)|(n-2)!$

Bill Dubuque
  • 272,048
emma funkhouser
  • 43
  • If $k$ is not a prime and can be written $k=ab$, can you prove that $k | (k-1)!$ ? – charmd Nov 20 '20 at 13:38
  • You basically need to prove that x^2 divides x!. Try to show that all three of x, x/2 and 2 appear in the factorial. Then show that the ratio of x! and x^2 results in an integer. – Ishraaq Parvez Nov 20 '20 at 13:40
  • Hint: $n-1=2\cdot (\frac12(n-1))$, and $2<\frac12(n-1)<n-2$. – TonyK Nov 20 '20 at 14:01
  • Beware that the answer you accepted is either incorrect or incomplete. – Bill Dubuque Nov 20 '20 at 21:51
  • 1
    If odd $,n\ge 7,$ then $,\color{#c00}{6\le }n!-!1 = \color{#c00}{2a}\mid a(a!+!1)(a!+!2)\mid (2a!-!1)!,$ by $,a!+!2\le 2a!-!1,$ by $,\color{#c00}{3\le a},,$ and this proof generalizes to any composite $,n\neq 4\ \ $ – Bill Dubuque Nov 20 '20 at 22:48

2 Answers2

1

This is not the same proof as before, I have changed the proof to the following:

If $n > 5$ is prime then $n-1$ is composite as there are no consecutive primes in this case. Hence, $n-1 = ab$ such that $1 < a < b < n-2 < n-1$ or $n-1 = p^{2}$ where $p$ is prime. For the first case this means that $a,b \in \{2,...n-2\}$ so that $ab|(n-2)!$ and $(n-1)|(n-2)!$. If $n-1 = p^{2}$, in order to get both $p$'s in the expansion of $(n-2)!$, we need to get $(n-2) \geq 2p = 2\sqrt{n-1}$ since if we consider the set $\{1,2,...,p,...2p\}$, this is the smallest set that when one takes the product, we have that $p^{2}$ divides the result. The previous inequality is exactly true for all $n>7$, but more specifically when $n$ prime. Therefore in either case, $(n-1)|(n-2)!$.

Derek Luna
  • 2,732
  • 8
  • 19
  • It might be to better to say "...and it is shown that either $n \leq 5$, or $n$ is composite instead of what I wrote. But that's getting pedantic. – Derek Luna Nov 20 '20 at 14:06
  • You seem to be implicitly assuming that if $,a\nmid b,$ then there is a prime $p$ dividing $a$ but not $b$. But this is false, e.g. consider $, b^2\nmid b,$ if $,b>1\ \ $ – Bill Dubuque Nov 20 '20 at 21:16
  • Does this only occur when of the form $b^{k} | b^{j}$ for $k > j$? I guess we see that $(n-1) \leq (n-2)!$ so what I write still holds I think right? – Derek Luna Nov 20 '20 at 21:20
  • You claim it is a proof by contrapositive, but what you proved is not the contrapositive of the OP's statement. – Bill Dubuque Nov 20 '20 at 21:27
  • If we have $(n-1) \nmid (n-2)!$, then $n \leq 5$ or $n$ is composite. This is what I've done. Since $(n-1) \nmid (n-2)!$, and it is known that $(n-1) \leq (n-2)!$, then we have the existence of such a prime. – Derek Luna Nov 20 '20 at 21:29
  • The argument in your answer is either incomplete or incorrect. I can't tell which is the case until you explain precisely why you believe it to be a proof of the contrapositive. – Bill Dubuque Nov 20 '20 at 21:34
  • (If all primes in the expansion of $a$ are in the expansion of $b$, and $a \leq b$, then $a|b) \implies$ (If $a \nmid b$, then either $a > b$, or there exists a prime in the expansion of $a$ that is not in the expansion of $b$). – Derek Luna Nov 20 '20 at 21:35
  • You appear to be attempting to either give another proof, or expand on the original proof. Please do that in your answer, not in the comments, so it is clear how it relates to the claims in the answer. – Bill Dubuque Nov 20 '20 at 21:37
  • I'm not sure if you are just being difficult or just trying to get me to write my proof better. – Derek Luna Nov 20 '20 at 21:38
  • I gave a counterexample above to what appears to be the inference you are using. Did you consider it carefully? – Bill Dubuque Nov 20 '20 at 21:39
  • I'm including that exponents are less in $a$ than in $b$ all throughout. – Derek Luna Nov 20 '20 at 21:40
  • Where in the answer did you "include" that, and where did you prove that it is it valid to do so here (obviously it is not valid in general as the counterexample shows). – Bill Dubuque Nov 20 '20 at 21:41
  • I see now that maybe the exponent could be larger somewhere but still $a \leq b$. I am getting confused at this point. If this proof has no saving just show me and I'll delete it. – Derek Luna Nov 20 '20 at 21:45
  • I can't answer that since you haven't answered my questions above, In particular, pecisely what is the "contrapositive" statement that you claim to be proving. I don't use chat, so please continue here if you wish to. – Bill Dubuque Nov 20 '20 at 21:49
  • Let us continue this discussion in chat. – Derek Luna Nov 20 '20 at 21:49
  • I see you edited it. Please explain why "we may suppose there exists..." – Bill Dubuque Nov 20 '20 at 22:02
  • I was talking to you on chat can you go there – Derek Luna Nov 20 '20 at 22:02
  • That if $(n-1) \nmid (n-2)!$, then $n$ is composite. Since the original statement is, if $n > 5$ and $n$ is prime, then $(n-1) | (n-2)!$. then since we already know that $(n-1) <= (n-2)!$, and by assumption $(n-1) \nmid (n-2)!$, [I thought we get a new prime, but this is not the case in general I see]. So my thinking is either there is a prime in$ (n-1)$ not in $(n-2)!$, or there is some prime power in $(n-1) $not in$ (n-2)!$ – Derek Luna Nov 20 '20 at 22:04
  • Which in the latter case I don't see how the argument can be salvaged. – Derek Luna Nov 20 '20 at 22:05
  • The platform won't let you delete until the OP unaccepts. If that doesn't occur you can flag it and ask a moderator to delete. Until then it would be courteous to prepend a remark explaining the issues, so that students are not misled (not everyone reads comments, esp. long comment threads). Best of luck. – Bill Dubuque Nov 20 '20 at 22:11
  • It seems like it fails. Could create some example probably where the case deduces to being only able to use the prime power $2^{4} = 16$. Then $p^{k} = 16=n-1$ so that $n = 17$ which is prime, not composite. You agree this can't be fixed right? – Derek Luna Nov 20 '20 at 22:11
  • @Bill Dubuque, but since we have $(n-1)\leq(n-2)!$, that means all prime powers in $(n-1)$ are also in $(n-2)!$, unless $n-1$ is a prime right $\implies n$ is composite since primes can't be consecutive in this case for $n > 5$. But then this looks like a direct proof. – Derek Luna Nov 20 '20 at 22:40
  • You shouldn't completely change the answer since the comments (and votes) are for the prior answer, so now everything is out of sync. Instead you should post the new answer separately. – Bill Dubuque Nov 21 '20 at 01:12
1

If $n$ is prime, then $n-1$ is composite unless $n=2 \text{ or } 3$. So the following line of argument begins by assuming $n>3$.

If $n>3$, then $n-1$ is composite, and it can be factored into two smaller integers, $(n-1)=ab$, each of which is smaller than $n-2$. This is true because an integer multiple of any positive integer $>1$ cannot equal the next larger integer (i.e. $k(n-2)\ne n-1$), so neither of $a,b$ can be as large as $(n-2)$.

If $n-1$ can be factored such that $a\ne b$, since $a,b<n-2$, they will each appear as separate terms in the product $(n-2)!$, and we are done. $a\mid (n-2)!$ and $b\mid (n-2)!$, so $ab\mid (n-2)!$, meaning $(n-1)\mid (n-2)!$. For odd primes, $n-1$ is even and can be always factored as $2$ times another number different from $2$, except when $n=5$. So far, the argument has been made assuming $n>3$, but now we must consider $5$ as a special case.

For $n=5$, where $n-1=4=2\times 2$ and $n-2=3$, the terms $1,2,3$ of $(n-2)!$ contain a factor of $2$ only once. In fact, $4\not \mid 3!=6$, so the proof fails for the case $n=5$, and we must require $n>5$ for it to be valid.

I hope that this not only shows that the proposition is true, but also explains why it must be conditioned on $n>5$

Keith Backman
  • 7,288