Greatest common divisor | Relatively prime

kingwinner

Claim: n! + 1 and (n+1)! + 1 are relatively prime.

How can we prove this? Can we use mathematical induction?
Base case: (n=1)
gcd(2,3)=1
Therefore, the statement is true for n=1.

Assuming the statment is true for n=k: gcd(k! + 1,(k+1)! + 1)=1 (induction hypothesis), we need to show that it's true for n=k+1. But I am stuck here. How can we use the induction hypothesis to prove this?

Or am I even on the right track thinking of using induction?

Thanks for any help!

CRGreathouse

Reduce mod k = n! + 1.

kingwinner

I have no background about modular arithmetic right now. It is an exercise from the first chapter of my book about "divisibility" and "relative primes". How can we prove it without assuming knowledge about modular arithmetic?

Petek

Try a proof by contradiction: Suppose that n! + 1 and (n + 1)! +1 are not relatively prime. Let p be a prime number that divides both numbers. Then p divides the difference, and so ... (Can you do the rest?). (Or if you haven't covered prime numbers yet, let d be any integer greater than 1 that divides both numbers and then continue as above.)

HTH

Petek

CRGreathouse

Write (n+1)! + 1 in terms of n! + 1.

kingwinner

I am not sure how to express (n+1)! + 1 in terms of n! + 1, the best I can think of is
(n+1)! + 1 = (n+1) n! +1

Proof by contradiction:
Suppose d is an integer greater than 1 that divides both n! + 1 and (n + 1)! +1
=> d divides (n + 1)! - n! = (n+1) n! -n! = n!(n+1-1) = (n!)n
i.e. d divides (n!)n
I am trying to reach a contradiction, but where is it??

CRGreathouse

Good first step: you've written it in terms of n!. Now you want to write it in terms of k = n! + 1, so replace all instances of n! by (k - 1) and expand.

kingwinner

OK!
(n+1)! + 1 = (n+1) n! +1

Let k= n! + 1 (=> n! = k-1)
Then (n+1)! + 1 = (n+1) n! +1 = (n+1)(k-1) +1
= nk -n +k -1 +1
=n (n! +1) - n + (n! +1)

But I still don't see where the "contradiction" is...

tiny-tim

Hi kingwinner!
ok, now subtract n! + 1 from that.

kingwinner

That's exactly what I did in post #6, but I can't find my contradiction...

tiny-tim

oh, sorry, I missed that …

ok, if a prime d divides the difference, then d | (n!)n, so d | n!, so …

kingwinner

Actually Petek was right, I haven't read the chapter about "primes" yet, so I assume no results about primes, so I let d to be any integer greater than 1 that divides both numbers. How should I continue?

But by the way, why if d is a prime and d | (n!)n => d | n! ???
One theorem in the later chapters says: If p is a prime, then p|(abc) => p|a OR p|b OR p|c.

Thanks!

tiny-tim

Yup … d | (n!)n => d | n! or d | n …

but if d | n, then d | n! anyway.

kingwinner

But I think the converse is false.
d | (n!)n => d|n or d|(n-1) or d|(n-2) or...or d|2 or d|1 or d|n
=> d|n or d|(n-1) or d|(n-2) or...or d|2 or d|1

tiny-tim

I agree …

but all you need is d | n!

kingwinner

Actually you are right. I don't know what I was thinking before.

d | (n!)n => d | n! or d | n
If d | n!, then d | n!
If d | n, then d | n!
So d | (n!)n => d | n! or d | n => d | n!

But we don't know what "n" is, so there does not seem to be any contradiction...?

tiny-tim

ah, but you started with …
so … ?