x Turn on thread page Beta
 You are Here: Home >< Maths

# Number theory - divisibility question watch

1. I have two consecutive numbers of a form , n is positive integer. I am looking for all prime numbers dividing both numbers. The answer is probably 13 for every n of a form 13k-7, where k is a positive integer but I cannot prove it.

I tried all sorts of things, nothing worked for me.

Can anyone help me out please with the proof, even sending me in the right direction would be greatly appreciated.
2. What do the \$ signs mean? They're confusing me.
3. Sorry, I was trying to use LATEX without reading the guide on how to use it.
4. The main things to use are that if p divides A and p divides B, then p divides A-B, A+B and AB. Use these rules a few times and it's not hard to show p divides 13.
5. I tried A-B and A+B before. Is it really necessary to go into AB (quartic expressions)?
6. Well let's assume there is a prime that divides both. Call this , which divides both and . Then divides the difference. Calculate the difference. That is all you need to do.
Have you heard of coprimality?
7. Hint: when you are calculating the difference, do not expand
8. (Original post by DerBoy)
Well let's assume there is a prime that divides both. Call this , which divides both and . Then divides the difference. Calculate the difference. That is all you need to do.
No it isn't. There are lots of primes that divide the difference. You need to narrow things down more than that.
9. (Original post by The_Special_2)
I tried A-B and A+B before. Is it really necessary to go into AB (quartic expressions)?
If A = (n+1)^2+3 and B = n^2+3, then looking at A-B gives you a "nice" condition relating p and n. But you need a 2nd condition to solve the problem.

Spoiler:
Show
I found it easiest to look at (A-B)^2 and compare with the expressions for A and B. There might be shorter methods, but this is pretty short, so it's certainly "good enough".
10. Thanks for the hint. If I do not expand, I get a difference of two squares of 2 consecutive numbers. There is probably something I should know about it but I don't. I still cannot se how I can deduce the answer. (13 and n=13k-7 if that is the answer)
11. Sorry, I didn't put enough thought into it. I thought since n^2 is coprime to (n+1)^2 that implied (n^2)+3 was coprime to ((n+1)^2)+3, apart from the case n=1. But I can't find a counterexample.
12. (Original post by DerBoy)
Sorry, I didn't put enough thought into it. I thought since n^2 is coprime to (n+1)^2 that implied (n^2)+3 was coprime to ((n+1)^2)+3, apart from the case n=1. But I can't find a counterexample.
13 divides 6^2+3 and 7^2+3.
13. (Original post by DFranklin)
If A = (n+1)^2+3 and B = n^2+3, then looking at A-B gives you a "nice" condition relating p and n. But you need a 2nd condition to solve the problem.

Spoiler:
Show
I found it easiest to look at (A-B)^2 and compare with the expressions for A and B. There might be shorter methods, but this is pretty short, so it's certainly "good enough".
Thanks for the hint. At the end I had to look at the "spoiler" but then everything became so easy. I spent more than 10 hours on something that could have been done in 10 minutes but that's life.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 28, 2010
Today on TSR

### Any tips for freshers...

who are introverts?

### More snow?!

Discussions on TSR

• Latest
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE