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 Post subject: Another Pretty Easy NT ProblemPosted: Mon, 9 Apr 2012 04:59:41 UTC
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There exist some nonzero integers such that:

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 ProblemPosted: Mon, 9 Apr 2012 22:40:43 UTC
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Joined: Tue, 16 Aug 2011 23:38:56 UTC
Posts: 124
rdj5933mile5math64 wrote:
There exist some nonzero integers such that:

is an integer.
Prove that abc is the cube of an integer.
Hint:
Spoiler:
Look at the primes.

If is an positive integer, then consider:

all positive integers, then:

with is an integer and is also an integer

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 Post subject: Re: Another Pretty Easy NT ProblemPosted: Mon, 9 Apr 2012 22:48:08 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12071
Location: Austin, TX
fam wrote:
rdj5933mile5math64 wrote:
There exist some nonzero integers such that:

is an integer.
Prove that abc is the cube of an integer.
Hint:
Spoiler:
Look at the primes.

If is an positive integer, then consider:

all positive integers, then:

with is an integer and 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 ProblemPosted: Tue, 10 Apr 2012 12:08:25 UTC
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Joined: Tue, 16 Aug 2011 23:38:56 UTC
Posts: 124
fam wrote:
rdj5933mile5math64 wrote:
There exist some nonzero integers such that:

is an integer.
Prove that abc is the cube of an integer.
Hint:
Spoiler:
Look at the primes.

If is an positive integer, then consider:

all positive integers, then:

with is an integer and 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 ProblemPosted: 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 is ___, hence either ... or ...

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 Post subject: Re: Another Pretty Easy NT ProblemPosted: Wed, 11 Apr 2012 02:54:40 UTC
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Joined: Wed, 4 Apr 2012 03:51:40 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 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

EDITs: Grammar Errors

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 Post subject: Re: Another Pretty Easy NT ProblemPosted: 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

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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 Post subject: Re: Another Pretty Easy NT ProblemPosted: 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

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 )

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 ProblemPosted: 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 )

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 ProblemPosted: Wed, 11 Apr 2012 19:16:39 UTC
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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.

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
EDIT2: I fail

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 Post subject: Re: Another Pretty Easy NT ProblemPosted: Thu, 12 Apr 2012 00:50:33 UTC
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rdj5933mile5math64 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.

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
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 ProblemPosted: Thu, 12 Apr 2012 02:14:44 UTC
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Joined: Wed, 4 Apr 2012 03:51:40 UTC
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Location: Hockeytown aka Detroit
rdj5933mile5math64 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.

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
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 be the highest power of that evenly divides (for any prime p, and any natural number x).
Rewrite the given expression as .
Note
Suppose is a multiple of some prime .
Wlog assume
We must have that .
Therefore,
I did more casework.
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 ProblemPosted: 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 or . The latter obviously doesn't give us an integer, so we must have . Now and the ultrametric property implies (since ). So and hence .

Note that this doesn't generalise to 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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 Post subject: Re: Another Pretty Easy NT ProblemPosted: Mon, 16 Apr 2012 18:24:42 UTC
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outermeasure wrote:
Spoiler:
Note that this doesn't generalise to 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 ProblemPosted: Tue, 17 Apr 2012 10:37:22 UTC
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rdj5933mile5math64 wrote:
outermeasure wrote:
Spoiler:
Note that this doesn't generalise to 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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