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**greg1313****Member**- Registered: 2016-12-19
- Posts: 17

Does the equation

have a solution in nonzero integers a, b, c ?

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The background would be intermediate/college algebra and number theory if

this could be proven or disproven.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Are a,b and c positive integers?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

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If are positive, then there are no solutions, since the AM-GM inequality would imply thatwhich is a contradiction. If the sign does not matter, then there are also no solutions. You can prove this by considering the case where one, two or three of are negative.

*Last edited by zetafunc (2016-12-19 22:20:05)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi thickhead;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**greg1313****Member**- Registered: 2016-12-19
- Posts: 17

bobbym wrote:

Hi;

Are a,b and c positive integers?

If I want a, b, c to be positive integers for other problems, I will state "positive integers" in those cases.

It seems some younger students aren't aware of negative integers, and I could see the clarification question

being asked here because of that.

Here, I included all integers, except zero.

So, it has been established that positive integers don't work, had that been the question. (I had already

researched that angle of the question some days prior.)

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greg1313 wrote:

If I want a, b, c to be positive integers for other problems, I will state "positive integers" in those cases.

It seems some younger students aren't aware of negative integers, and I could see the clarification question

being asked here because of that.Here, I included all integers, except zero.

So, it has been established that positive integers don't work, had that been the question. (I had already

researched that angle of the question some days prior.)

You can show that there aren't any solutions by considering the cases where one, two, or three of (a,b,c) are negative.

If a, b, c are all negative, then it's easy to show that a < b < c < a, which is impossible.

If a, b are negative but c is positive, then you can also establish that a < b < c. Now multiply your original expression by abc and derive a contradiction by proving the LHS is negative, whilst the RHS is positive.

I'll leave the final case to you as an exercise.

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**greg1313****Member**- Registered: 2016-12-19
- Posts: 17

zetafunc wrote:

greg1313 wrote:Here, I included all integers, except zero.

You can show that there aren't any solutions by considering the cases where one, two, or three of (a,b,c) are negative.

If a, b are negative but c is positive, then you can also establish that a < b < c. Now multiply your original expression

by abc and derive a contradiction by proving the LHS is negative, whilst the RHS is positive.

I isolated your second case to further discuss.

My work:

----------

Let r, p be positive integers.

Suppose a = -r

Suppose b = -p

Keep c positive.

Multiply each side by rpc:

Add two terms to each side:

I don't see any contradiction of signs here, because the LHS is positive and so is the RHS.

It just means this case is inconclusive.

If your third case is based on the technique of the second case, then that might be inconclusive, too.

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greg1313 wrote:

Add two terms to each side:

I don't see any contradiction of signs here, because the LHS is positive and so is the RHS.

It just means this case is inconclusive.

That isn't what I advised. Go back to your previous expression and prove that the LHS is negative. The point I was trying to make was that you'll need to establish an ordering relation between p, c, r to do it, as per the below:

zetafunc wrote:

If a, b are negative but c is positive, then you can also establish that a < b < c. Now multiply your original expression by abc and derive a contradiction by proving the LHS is negative, whilst the RHS is positive.

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

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**greg1313****Member**- Registered: 2016-12-19
- Posts: 17

zetafunc wrote:

Go back to your previous expression and prove that the LHS is negative. The point I was trying to make was that you'll need to establish an ordering relation between p, c, r to do it, as per the below:

zetafunc wrote:

You must have an error somewhere. I know I am asking about all integers, except zero.

But it should not make a difference if I were to ask about a, b being negative, c is positive and all three belonging

to the set of real numbers. And, I will make sure [in general, and in my example below] that a < b < c.

I came up with this counterexample for you:

If you work this out, you will see that the expression on the left-hand side simplifies to 1.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

In general where a and b negative and a < b:

Let r, p be positive integers

Let b = -r

To make sure a < b,

let a = -p - r

c = c

Substitute:

The sides do not contradict each other in signs.

If I were to multiply each side by cp(p + r), I would get:

You're not going to manipulate that and get a negative LHS versus a positive RHS.

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