Another Pretty Easy NT Problem

rdj5933mile5math64 · **Posted:** Mon, 9 Apr 2012 04:59:41 UTC

There exist some nonzero integers such that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$
is an integer.
Prove that abc is the cube of an integer.
Hint:

Spoiler:

fam · **Joined:** Tue, 16 Aug 2011 23:38:56 UTC **Posts:** 124

rdj5933mile5math64 wrote:

There exist some nonzero integers such that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$
is an integer.
Prove that abc is the cube of an integer.
Hint:

Spoiler:

If

is an positive integer, then consider:

a = n ; b = 2n ; c = 4n

all positive integers, then:

$a \times b \times c = 8n^3$

(2n)^3= 8n^3

with

is an integer and $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}= 5$ is also an integer

Shadow · **Posted:** Mon, 9 Apr 2012 22:48:08 UTC

fam wrote:

rdj5933mile5math64 wrote:

There exist some nonzero integers such that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$
is an integer.
Prove that abc is the cube of an integer.
Hint:

Spoiler:

If

is an positive integer, then consider:

a = n ; b = 2n ; c = 4n

all positive integers, then:

$a \times b \times c = 8n^3$

(2n)^3= 8n^3

with

is an integer and $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}= 5$ is also an integer

This doesn't solve the problem. The op wants you to show that abc MUST be the cube of an integer, you cannot just set values at the beginning.

fam · **Joined:** Tue, 16 Aug 2011 23:38:56 UTC **Posts:** 124

Shadow wrote:

fam wrote:

rdj5933mile5math64 wrote:

There exist some nonzero integers such that:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$
is an integer.
Prove that abc is the cube of an integer.
Hint:

Spoiler:

If

is an positive integer, then consider:

a = n ; b = 2n ; c = 4n

all positive integers, then:

$a \times b \times c = 8n^3$

(2n)^3= 8n^3

with

is an integer and $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}= 5$ is also an integer

This doesn't solve the problem. The op wants you to show that abc MUST be the cube of an integer, you cannot just set values at the beginning.

What method of proof do you suggest?

outermeasure · **Posted:** Tue, 10 Apr 2012 13:51:36 UTC

fam wrote:

What method of proof do you suggest?

Certainly your alleged "proof" is not a proof. For a start, there are integers beside 5 which you can expressed as a/b+b/c+c/a for some nonzero integers a,b,c (I'll point you to the bottom of this page, for example).

The probably easiest proof starts with: We can assume, without loss of generality, that $\mathop{\mathrm{gcd}}(a,b,c)$ is ___, hence either ... or ...

rdj5933mile5math64 · **Posted:** Wed, 11 Apr 2012 02:54:40 UTC

outermeasure wrote:

this page

uhmmmmm darn I need to get smarter at math... what is this and where could I find more information on this? (As much as I try, I can't learn much math beyond bc calculus by reading random arxiv papers

)

outermeasure wrote:

The probably easiest proof starts with: We can assume, without loss of generality, that $\mathop{\mathrm{gcd}}(a,b,c)$ is ___, hence either ... or ...

I think that this is the easiest proof (assuming of course you are thinking what I am thinking), but obviously I can't be sure. That's the right first step. I probably should have included that in my hint.

Spoiler:

EDITs: Grammar Errors

outermeasure · **Posted:** Wed, 11 Apr 2012 14:11:26 UTC

rdj5933mile5math64 wrote:

uhmmmmm darn I need to get smarter at math... what is this and where could I find more information on this? (As much as I try, I can't learn much math beyond bc calculus by reading random arxiv papers

)

This is a handy reference for some computer calculation on some family of elliptic curves. Basically there is no need to understand any of these for this problem, I was just pointing out there are many more integers that can be written as a/b+b/c+c/a for rational (or integral, since we can clear denominators) a,b,c than just the number 5. And while you can probably understand some mathematics of this if you read around (e.g. the many introductory texts on elliptic curve cryptography for computer scientists include very short crash course(s) on basic number theory of elliptic curves accessible to people without mathematics background), I can't think of a single book that would go from school mathematics to the more advanced theory that goes into this calculation.

rdj5933mile5math64 wrote:

Spoiler:

Yes, that is the first step. Then

Spoiler:

rdj5933mile5math64 · **Posted:** Wed, 11 Apr 2012 16:41:58 UTC

outermeasure wrote:

And while you can probably understand some mathematics of this if you read around (e.g. the many introductory texts on elliptic curve cryptography for computer scientists include very short crash course(s) on basic number theory of elliptic curves accessible to people without mathematics background), I can't think of a single book that would go from school mathematics to the more advanced theory that goes into this calculation. I can't think of a single book that would go from school mathematics to the more advanced theory that goes into this calculation.

I know basic NT like UFD's. I just don't know topology yet

outermeasure wrote:

rdj5933mile5math64 wrote:

Spoiler:

Yes, that is the first step. Then

Spoiler:

lolz I already solved the problem. Hmm I actually did it differently saying:

Spoiler:

I posted it in hopes of maybe getting a way to find a more general result. In particular, it looks like if x1/x2 +x2/x3+x3/4+....xn/x1 is an integer then x1x2....xn is a perfect n power.

Shadow · **Posted:** Wed, 11 Apr 2012 17:17:26 UTC

rdj5933mile5math64 wrote:

Spoiler:

I posted it in hopes of maybe getting a way to find a more general result. In particular, it looks like if x1/x2 +x2/x3+x3/4+....xn/x1 is an integer then x1x2....xn is a perfect n power.

As for the hint you post, you're to assume that the mutual gcd is 1, so that hint is wrong.

As for "more general results" that's not a topic for a proposed problem, that's a topic of inquiry. These are specific problems with specific solutions, not theorem proving.

rdj5933mile5math64 · **Posted:** Wed, 11 Apr 2012 19:16:39 UTC

Shadow wrote:

As for the hint you post, you're to assume that the mutual gcd is 1, so that hint is wrong.

Nope it's right. Suppose gcd(x1,x2,x3,....,xn)=k, we note that the desired result (if x1/x2 +x2/x3+x3/4+....xn/x1 is an integer, then x1x2....xn is a perfect n power) follows iff the result is true for y1=x1/k, y2=x2/k,.... yn=xn/k.

Shadow wrote:

As for "more general results" that's not a topic for a proposed problem, that's a topic of inquiry. These are specific problems with specific solutions, not theorem proving.

Actually outermeasure's method generalizes well. I should have noticed how well that works earlier. fail :!:

EDIT: darn forgot the second part of my post :oops:

EDIT2: I fail

Shadow · **Posted:** Thu, 12 Apr 2012 00:50:33 UTC

rdj5933mile5math64 wrote:

Shadow wrote:

As for the hint you post, you're to assume that the mutual gcd is 1, so that hint is wrong.

Nope it's right. Suppose gcd(x1,x2,x3,....,xn)=k, we note that the desired result (if x1/x2 +x2/x3+x3/4+....xn/x1 is an integer, then x1x2....xn is a perfect n power) follows iff the result is true for y1=x1/k, y2=x2/k,.... yn=xn/k.

Shadow wrote:

As for "more general results" that's not a topic for a proposed problem, that's a topic of inquiry. These are specific problems with specific solutions, not theorem proving.

Actually outermeasure's method generalizes well. I should have noticed how well that works earlier. fail :!:

EDIT: darn forgot the second part of my post :oops:

EDIT2: I fail

Notice if a,b,c all share a common factor, then it doesn't change your ratio sum, so there's no point in not assuming gcd(a,b,c)=1, that's why the hint doesn't help.

rdj5933mile5math64 · **Posted:** Thu, 12 Apr 2012 02:14:44 UTC

Shadow wrote:

rdj5933mile5math64 wrote:

Shadow wrote:

As for the hint you post, you're to assume that the mutual gcd is 1, so that hint is wrong.

Nope it's right. Suppose gcd(x1,x2,x3,....,xn)=k, we note that the desired result (if x1/x2 +x2/x3+x3/4+....xn/x1 is an integer, then x1x2....xn is a perfect n power) follows iff the result is true for y1=x1/k, y2=x2/k,.... yn=xn/k.

Shadow wrote:

As for "more general results" that's not a topic for a proposed problem, that's a topic of inquiry. These are specific problems with specific solutions, not theorem proving.

Actually outermeasure's method generalizes well. I should have noticed how well that works earlier. fail :!:

EDIT: darn forgot the second part of my post :oops:

EDIT2: I fail

Notice if a,b,c all share a common factor, then it doesn't change your ratio sum, so there's no point in not assuming gcd(a,b,c)=1, that's why the hint doesn't help.

The hint was originally for my solution, and it is used for simplifying different cases. I haven't really gotten a chance to work out outermeasure's solution entirely, but I think that it is much better than mine.

The gcd thing is used mostly to help with the cases. In particular,

Spoiler:

My Solution:

Spoiler:

outermeasure · **Posted:** Thu, 12 Apr 2012 04:33:50 UTC

My solution is slightly different:

Spoiler:

rdj5933mile5math64 · **Posted:** Mon, 16 Apr 2012 18:24:42 UTC

outermeasure wrote:

Spoiler:

hmmm I thought there was a different solution, but after trying to finish this thought process out, things sort of fell apart lolz...

But I don't understand the application of the last bit, about n being even implying that it doesn't divide n(n-1)/2. What does that mean in terms of this problem?

outermeasure · **Posted:** Tue, 17 Apr 2012 10:37:22 UTC

rdj5933mile5math64 wrote:

outermeasure wrote:

Spoiler:

hmmm I thought there was a different solution, but after trying to finish this thought process out, things sort of fell apart lolz...

But I don't understand the application of the last bit, about n being even implying that it doesn't divide n(n-1)/2. What does that mean in terms of this problem?

In this problem, you had the exact powers of p dividing your a,b,c are 0,m,2m -- even if (and that is a big if to ask for, in fact I don't believe it is true) this generalises to 0,m,2m,3m,....,(n-1)m, unless m is even you won't get n dividing 0+m+...+(n-1)m=mn(n-1)/2 when n is even.

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