# Greatest common divisor | Relatively prime

1. Jan 7, 2010

### 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!

2. Jan 7, 2010

### CRGreathouse

Reduce mod k = n! + 1.

3. Jan 7, 2010

### 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?

Last edited: Jan 7, 2010
4. Jan 7, 2010

### 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

5. Jan 7, 2010

### CRGreathouse

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

6. Jan 7, 2010

### 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

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??

Last edited: Jan 8, 2010
7. Jan 8, 2010

### 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.

8. Jan 9, 2010

### 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...

9. Jan 9, 2010

### tiny-tim

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

10. Jan 9, 2010

### kingwinner

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

11. Jan 9, 2010

### tiny-tim

oh, sorry, I missed that

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

12. Jan 9, 2010

### 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!

13. Jan 9, 2010

### tiny-tim

Yup … d | (n!)n => d | n! or d | n …

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

14. Jan 9, 2010

### 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

15. Jan 9, 2010

### tiny-tim

I agree

but all you need is d | n!

16. Jan 9, 2010

### 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...?

17. Jan 9, 2010

### tiny-tim

ah, but you started with …
so … ?

18. Jan 9, 2010

### 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?

19. Jan 9, 2010

### 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

20. Jan 9, 2010

### tiny-tim

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? )

21. Jan 9, 2010

### kingwinner

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

Thanks!

22. Jan 9, 2010

### tiny-tim

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

23. Jan 9, 2010

### 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

Last edited: Jan 9, 2010
24. Jan 9, 2010

### Petek

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

Petek

25. Jan 9, 2010

### kingwinner

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

Last edited: Jan 9, 2010