Proof: a/b + b/a is Int iff a=b

[DrAlexMV]

Homework Statement

Prove: a/b + b/a equals an integer, for any a,b nonzero positive integers, if and only if a = b

Homework Equations

Divisibility: d|e implies e = dj where j is an element of j for d and e element of the integers.

The Attempt at a Solution

1. Let a/b + b/a = k where k is an element of the integers

2. a/b + b/a = (a^2 + b^2) / ab

3. (a^2 + b^2) = kab

4. Let (a^2 + b^2) = c where c is an integer

5. c = kab

6. Step five implies the three following things: a|c , b|c , ab|c

7. If a|c and b|c, then ab does not divide c and therefore step 6 demonstrates a contradiction.

8. Therefore a must equal b

Is this a solid proof? If not, could you tell me where the weakness is?

EDIT : If a and b are relatively prime. I cannot use my contradiction then! Any ideas ?

Thank you for your time.

 

[mtayab1994]

One thing that positive integer could be odd or even.

 

[DrAlexMV]

Eh what if a and b are relatively prime. I cannot use my contradiction then! Any ideas ?

 

[DrAlexMV]

mtayab1994, what do you mean with that?

 

[mtayab1994]

sorry my earlier comment was a typo i wanted to say if they were prime like you just stated

 

[DrAlexMV]

Yeah, I thought about that after I typed it. Any idea on how to solve this?

 

[mtayab1994]

Well it seems that it really doesn't matter if they're prime, because take for example 5 ( a prime number). In this case we get c=a^2+b^2=50 so c/a=10 and c/b=10 and c/ab=2; and this will work for any number and that number c divided by ab is the same as a/b+b/a. Hopefully you can something in this.

 

[Office_Shredder]

I don't understand the contradiction. If a and b divide c ab can definitely divide c as well. In fact I don't see anything about a=b until after you got a contradiction already.

At step 3 just consider the equation mod a and b

 

[mtayab1994]

I don't understand the contradiction. If a and b divide c ab can definitely divide c as well

Yes that's what i tried showing him in the previous reply.

 

[Dick]

I'm a little vague on what you think the contradiction is. I would assume a and b are relatively prime to begin with. (Otherwise you could just remove the common factors). Now suppose p is a prime divisor of a. What could you conclude from (a^2 + b^2) = kab?

 

[DrAlexMV]

Relatively prime does not mean that they are prime. Relatively prime means that the gcd(a,b) = 1. They have no common factors. My contradiction fails if a and b have no common factors.

Office Shredder, if a|c and b|c, then ab|c if gcd(a,b) = 1. Otherwise it does not work.

 

[DrAlexMV]

Would that mean that p|(a^2 + b^2) ?

 

[mtayab1994]

Would that mean that p|(a^2 + b^2) ?

Yes.

 

[DrAlexMV]

Where does that take me?

 

[Dick]

Would that mean that p|(a^2 + b^2) ?

If there are multiple people responding it's a good idea to hit the Quote button before you reply to make it clear who you are responding to. But yes, assume gcd(a,b)=1 and try to derive a contradiction with that.

 

[mtayab1994]

I'm not so sure.

 

[DrAlexMV]

If there are multiple people responding it's a good idea to hit the Quote button before you reply to make it clear who you are responding to. But yes, assume gcd(a,b)=1 and try to derive a contradiction with that.

So,
1. a^2 + b^2 = kab

2. Let gcd(a,b) = 1

3. Let p be a prime divisor of a

4. Then p|a^2 + b^2

...

I'm thinking but it is not clicking.

 

[Dick]

So,
1. a^2 + b^2 = kab

2. Let gcd(a,b) = 1

3. Let p be a prime divisor of a

4. Then p|a^2 + b^2

...

I'm thinking but it is not clicking.

Can't you show p must divide b as well?

 

[DrAlexMV]

Can't you show p must divide b as well?

Would this work?

a^2 + b^2 = pj where j is an element of the integers
a = pw where w is an element of the integers

Then:
p^2w^2 + b^2 = pj

we reorganize to:
b^2 = p(j - pw^2)

we let j - pw^2 = v where v is an element of the integers

so b^2 = pv

that implies p|b^2 and b times b will not create a factor p since p is prime therefore,

p|b

 

[Dick]

Would this work?

a^2 + b^2 = pj where j is an element of the integers
a = pw where w is an element of the integers

Then:
p^2w^2 + b^2 = pj

we reorganize to:
b^2 = p(j - pw^2)

we let j - pw^2 = v where v is an element of the integers

so b^2 = pv

that implies p|b^2 and b times b will not create a factor p since p is prime therefore,

p|b

That's a little more complicated than it needs to be but if p|a then p|a^2 and p|kab, so p|b^2. And if p is prime, then sure, p|b. Doesn't that contradict gcd(a,b)=1?

 

[DrAlexMV]

Yes, I saw that whenever you told me to prove that p|b. Thank you a lot, this proof really made me think for some reason.