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 Post subject: Another Pretty Easy NT Problem
PostPosted: Mon, 9 Apr 2012 04:59:41 UTC 
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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:
Look at the primes.

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Mon, 9 Apr 2012 22:40:43 UTC 
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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:
Look at the primes.


If n 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 2n is an integer and \frac{a}{b}+\frac{b}{c}+\frac{c}{a}= 5 is also an integer


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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Mon, 9 Apr 2012 22:48:08 UTC 
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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:
Look at the primes.


If n 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 2n 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.

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Tue, 10 Apr 2012 12:08:25 UTC 
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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:
Look at the primes.


If n 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 2n 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?


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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Tue, 10 Apr 2012 13:51:36 UTC 
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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 ...

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Wed, 11 Apr 2012 02:54:40 UTC 
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outermeasure 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 :( )

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:
wlog \mathop{\mathrm{gcd}}(a,b,c)=1


EDITs: Grammar Errors

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Wed, 11 Apr 2012 14:11:26 UTC 
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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:
wlog \mathop{\mathrm{gcd}}(a,b,c)=1


Yes, that is the first step. Then
Spoiler:
any prime p dividing abc must only divide one or two of a,b,c. Reject the former, and ...

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Wed, 11 Apr 2012 16:41:58 UTC 
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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:
wlog \mathop{\mathrm{gcd}}(a,b,c)=1


Yes, that is the first step. Then
Spoiler:
any prime p dividing abc must only divide one or two of a,b,c. Reject the former, and ...


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

Spoiler:
Let p be any prime that divides a. Then, p|b, p|c. From there the result follows. (Hint: use v_p)

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.

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Wed, 11 Apr 2012 17:17:26 UTC 
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rdj5933mile5math64 wrote:
Spoiler:
Let p be any prime that divides a. Then, p|b, p|c. From there the result follows. (Hint: use v_p)

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.

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Wed, 11 Apr 2012 19:16:39 UTC 
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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

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Thu, 12 Apr 2012 00:50:33 UTC 
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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.

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Thu, 12 Apr 2012 02:14:44 UTC 
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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:
by assuming this we don't have to deal with p|b and p|c


My Solution:
Spoiler:
Wlog let gcd(a,b,c)=1. Also, let v_p(x) be the highest power of p that evenly divides x (for any prime p, and any natural number x).
Rewrite the given expression as \frac{a^2c+b^2a+c^2b}{abc}.
Note a|c^2b, : b|a^2c, : c|b^2a
Suppose a is a multiple of some prime p.
Wlog assume p|b
We must have that v_p(b^2a) \ge v_p(abc).
Therefore, v_p(a^2+cb) \ge v_p(ab)
I did more casework. v_p(b)<2v_p(a), v_p(b)>2v_p(a), v_p(b)=2v_p(a)
The only working case is the last.
p was arbitrary. so that's that

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Thu, 12 Apr 2012 04:33:50 UTC 
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My solution is slightly different:
Spoiler:
Continuing from the last hint:

Cyclic permuting a,b,c if necessary, we may assume 0=v_p(a)<v_p(b)\leq v_p(c) or 0=v_p(c)<v_p(b)\leq v_p(a). The latter obviously doesn't give us an integer, so we must have 0=v_p(a)<v_p(b)\leq v_p(c). Now v_p(\frac{a}{b}+\frac{b}{c})\geq \inf\{v_p(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}),v_p(c)\}\geq 0 and the ultrametric property implies v_p(\frac{a}{b})=v_p(\frac{b}{c}) (since v_p(\frac{a}{b})<0). So v_p(c)=2v_p(b) and hence v_p(abc)=3v_p(b).

Note that this doesn't generalise to \frac{x_1}{x_2}+\frac{x_2}{x_3}+\cdots+\frac{x_n}{x_1} though --- a priori there is no reason why only one of them has v_p zero, and also there is no reason to expect the jumps must be evenly spaced. Moreover, when n is even, it does not divide n(n-1)/2.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Mon, 16 Apr 2012 18:24:42 UTC 
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outermeasure wrote:
Spoiler:
Note that this doesn't generalise to \frac{x_1}{x_2}+\frac{x_2}{x_3}+\cdots+\frac{x_n}{x_1} though --- a priori there is no reason why only one of them has v_p zero, and also there is no reason to expect the jumps must be evenly spaced. Moreover, when n is even, it does not divide n(n-1)/2.


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?

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 Post subject: Re: Another Pretty Easy NT Problem
PostPosted: Tue, 17 Apr 2012 10:37:22 UTC 
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rdj5933mile5math64 wrote:
outermeasure wrote:
Spoiler:
Note that this doesn't generalise to \frac{x_1}{x_2}+\frac{x_2}{x_3}+\cdots+\frac{x_n}{x_1} though --- a priori there is no reason why only one of them has v_p zero, and also there is no reason to expect the jumps must be evenly spaced. Moreover, when n is even, it does not divide n(n-1)/2.


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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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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