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 statement 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]

How can we prove this?

Reduce mod k = n! + 1.

 

[kingwinner]

Reduce mod k = n! + 1.

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]

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?

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

 

[kingwinner]

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

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

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]

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

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]

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.

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!

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

ok, now subtract n! + 1 from that.

 

[kingwinner]

Hi kingwinner!


ok, now subtract n! + 1 from that.

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

 

[tiny-tim]

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

oh, sorry, I missed that …

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

 

[kingwinner]

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

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]

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.

Yup … d | (n!)n => d | n! or d | n …

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

 

[kingwinner]

Yup … d | (n!)n => d | n! or d | n …

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

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]

… d | n!

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

ah, but you started with …

Proof by contradiction:
Suppose d is an integer greater than 1 that divides both n! + 1 and (n + 1)! +1

so … ?

 

[kingwinner]

The claim is n! + 1 and (n+1)! + 1 are relatively prime for ALL n E N.

So to use proof by contradiction, we assume that
there exists SOME n E N such that n! + 1 and (n+1)! + 1 are not relatively prime.

But we don't know what that "n" is. It could be 1000, or 10205.

d | n! => n! = d z for some z E Z where d is a prime. I don't see anything contradictory with this. Prime factors are allowed in the factorization, right?

 

[Petek]

If you have shown that d|n!, then the contradiction arises from the assumption that d also divides n! + 1 (since consecutive integers are relatively prime). However, it does not follow that d|(n)(n!) implies that d|n!. For example 32|(6)(6!), but 32 does not divide 6!.

This problem is trickier than I thought when I gave the above hint.

Petek

 

[tiny-tim]

The claim is n! + 1 and (n+1)! + 1 are relatively prime for ALL n E N.

So to use proof by contradiction, we assume that
there exists SOME n E N such that n! + 1 and (n+1)! + 1 are not relatively prime.

Yes, and that means there is a prime d dividing both of them,

and you proved that that d divides n!.

So you have d | n! + 1 and (n+1)! + 1 and n!

(if you still haven't got it … if n = 4, can d divide both 24 and 25? )

 

[kingwinner]

If you have shown that d|n!, then the contradiction arises from the assumption that d also divides n! + 1 (since consecutive integers are relatively prime). However, it does not follow that d|(n)(n!) implies that d|n!. For example 32|(6)(6!), but 32 does not divide 6!.

This problem is trickier than I thought when I gave the above hint.

Petek

OK, I follow your arguments that IF d|n!, then we're done.

But as you said "it does not follow that d|(n)(n!) implies that d|n!", then what should I do instead?

Thanks!

 

[tiny-tim]

… However, it does not follow that d|(n)(n!) implies that d|n!. For example 32|(6)(6!), but 32 does not divide 6!.

But as you said "it does not follow that d|(n)(n!) implies that d|n!", then what should I do instead?

If d is prime, d|(n)(n!) does imply that d|n!

 

[Petek]

OK, I think that I now have a valid solution. Here're some hints:

Use the Euclidean algorithm to derive the following identity:

(n + 1)! + 1 = (n - 1)(n! + 1) + (n! + 1 - n)

(or just simplify the RHS of the equation to verify the equality).

Conclude that if d divides both (n + 1)! + 1 and n! + 1, then d divides n! +1 - n. Using the techniques discussed in this thread, show that this implies that d divides n. It's already been shown above that d|n gives a contradiction.

HTH

Petek

 

[Petek]

If d is prime, d|(n)(n!) does imply that d|n!

Correct, but the OP hadn't covered prime numbers when solving this problem.

Petek

 

[kingwinner]

OK, I think that I now have a valid solution. Here're some hints:

Use the Euclidean algorithm to derive the following identity:

(n + 1)! + 1 = (n - 1)(n! + 1) + (n! + 1 - n)

(or just simplify the RHS of the equation to verify the equality).

Conclude that if d divides both (n + 1)! + 1 and n! + 1, then d divides n! +1 - n. Using the techniques discussed in this thread, show that this implies that d divides n. It's already been shown above that d|n gives a contradiction.

HTH

Petek

I haven't learned Euclidean algorithm yet...and also, I fail to see why (n + 1)! + 1 = (n - 1)(n! + 1) + (n! + 1 - n). I tried simplifying the RHS but I got n! n instead of (n + 1)! + 1.

And also, to get a contradiction, I think we need d|(n!) instead of d|n

 

[tiny-tim]

I haven't learned Euclidean algorithm yet...and also, I fail to see why (n + 1)! + 1 = (n - 1)(n! + 1) + (n! + 1 - n). I tried simplifying the RHS but I got n! n instead of (n + 1)! + 1.

kingwinner, why are you bothering with this? The equation you found yourself …

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

… is much simpler.

 

[Petek]

I haven't learned Euclidean algorithm yet...and also, I fail to see why (n + 1)! + 1 = (n - 1)(n! + 1) + (n! + 1 - n). I tried simplifying the RHS but I got n! n instead of (n + 1)! + 1.

And also, to get a contradiction, I think we need d|(n!) instead of d|n

Sorry about that. The equation should be

(n + 1)! + 1 = n(n! + 1) + (n! + 1 - n)

and everything else I wrote is OK (I hope). I just mentioned the Euclidean algorithm to motivate how I derived the equation. Finally, it was already pointed out that d|n implies d|n!.

Petek

 

[Petek]

kingwinner, why are you bothering with this? The equation you found yourself …

… is much simpler.

@tiny-tim,

The difficulty is deriving a contradiction from d|n(n!) without using the theorem that p|ab implies p|a or p|b, with p prime. It's easy if you can use that theorem, but seems to require additional reasoning if you can't. Or perhaps you do have an easier argument?

Petek

 

[kingwinner]

Sorry about that. The equation should be

(n + 1)! + 1 = n(n! + 1) + (n! + 1 - n)

and everything else I wrote is OK (I hope). I just mentioned the Euclidean algorithm to motivate how I derived the equation. Finally, it was already pointed out that d|n implies d|n!.

Petek

(n+1)! + 1 = n (n! +1) - n + (n! +1) = (n+1)(n! +1) - n
=> n = (n+1)[n! +1] - [(n+1)! + 1]
By assumption, d divides both (n + 1)! + 1 and n! + 1
=> d divides any integer linear combination of (n + 1)! + 1 and n! + 1
=> d divides (n+1)[n! +1] - [(n+1)! + 1]
=> d|n
=> d|(n!) (since a|b => a|(bc) for any c E Z and n! = n (n-1)! )
So d|(n!) and d|(n! +1) where d>1 <-----contradiction
Thus, no integer greater than 1 divides both n! + 1 and (n + 1)! +1.


(so I think the key point is to express "n" as a linear combiation of (n + 1)! + 1 and n! + 1)

Are my reasonings here correct?

 

[Petek]

Looks good, but let me review the details tomorrow.

Petek