# Prove m + m/1

1. May 20, 2007

### disregardthat

1. The problem statement, all variables and given/known data

Prove that if m is a positive rational number, then:

m + 1/m = integer

Only when m = 1

2. Relevant equations

Don't know any

3. The attempt at a solution

That's the problem, I don't know where to start. I have tryed a few things, but none of them works out for me.

I don't know much of proofs and such, the problem I have is always where shall I start? Do you have any tips?

EDIT: The topics name is not correct, the correct question is in the topic

2. May 20, 2007

If m is a positive rational number, then m = p/q, for some natural numbers p and q. Hence, you have p/q + q/p = integer => p/q & q/p are integers. Now, what does that mean?

3. May 20, 2007

### Staff: Mentor

Just because the sum of two numbers is an integer does not mean the two numbers themselves are integers. This final step needs further justification.

4. May 20, 2007

### disregardthat

I was at that stage:

a/b + b/a = integer

But I couldn't state that this only works when m=1

Is it ok to say: a and b is integers, if a is more that 1, we would have a decimal number in b/a and an integer in a/b, if b is more than one, we would have a decimal number in a/b and an integer in b/a.

Since m must be positive, and that we can't divide by 0, m must be 1 to get an integer.

Last edited: May 20, 2007
5. May 20, 2007

### Staff: Mentor

How would you express a/b+b/a as a fraction? What are the conditions that this fraction be an integer?

Now put this back in the context of the original problem, which is to prove m+1/m is an integer only if m is 1.

6. May 20, 2007

a/b + b/a is not an integer if a doesn't divide b or if b doesn't divide a (or both). So, for a/b + b/a to be an integer, a must divide b, and b must divide a.

7. May 20, 2007

### Staff: Mentor

radou, you just gave out the answer. Not a good idea in the homework section ...

8. May 20, 2007

Hm, I'd let Jarle decide about that.

9. May 20, 2007

### Dick

This is all true, except that the sum of two nonintegers can be a integer. So it's not so clear and I don't think the answer is given away yet. You'll want to clear the fractions in a/b+b/a=1 and assume a and b are relatively prime.

Last edited: May 20, 2007
10. May 21, 2007

### disregardthat

I don't know how to prove it, I tryed, but I can't

Last edited: May 21, 2007
11. May 21, 2007

### Dick

I'm not sure what you mean by 'shortened'. I would write it as a^2+b^2=a*b. Is this relation possible if a and b have no common divisors?

12. May 21, 2007

### HallsofIvy

Staff Emeritus
I'm a bit unclear as to whether the problem is to show that m+ 1/m is an integer or (m+1)/m is an integer only if m= 1.

If it is the first case, I don't think you need to worry about "divisors". Since the difference of two integers is an integer, if m+ 1/m= n, an integer, then 1/m= n- m is an integer. What does that tell you?

13. May 21, 2007

### Staff: Mentor

Halls, it looks like you assumed that m is an integer. It is a rational.

14. May 21, 2007

### Dick

You can assume a and b have no common divisors, right? Otherwise you could just simplify the fraction. This mean one of a or b has a prime divisor that does not divide the other. Suppose p divides a but not b. How many of the terms in a^2+b^2=a*b are divisible by p? How many are not? Is there anything wrong here?

Last edited: May 22, 2007
15. May 21, 2007

### Staff: Mentor

Dick, you arrived at $a/b+b/a=1$ in post #9 and then used that in posts #11 and #14. However, you do not know $a/b+b/a=1$. The given condition is that $a/b+b/a=n$, where $a, b, n \in \mathbb N$. So the problem becomes one of showing that $a^2+b^2=nab$ only has solution if $a=b=1$.

16. May 21, 2007

### Dick

Ooooops. Sorry! Same argument, though.

17. May 21, 2007

### Jimmy Snyder

That should read "only has solution if $a=b$"

Last edited: May 21, 2007
18. May 21, 2007

Nah, it should read "only has solution in the set of integers".

19. May 21, 2007

### Jimmy Snyder

a and b may both be integers and yet not be a solution. a = b is crucial. a and b were chosen from the set of integers.

20. May 21, 2007