Math Is Fun Forum

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Number theory #1

I have an idea. Let this topic be the number theory QA topic.

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krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Number theory #1

"e" means "element"
N means Natural(>=1)
Z means Integer
Sq means Perfect square set:
x e Sq means "x is perfect square"
P means prime number set.

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krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Number theory #1

NumbTh1:
If a,b ∈ N and ab≡0(mod a+b) prove that a²b²/(a+b) has a rom x²y³, where x,y ∈ N

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ryos

Member

Registered: 2005-08-04

Posts: 394

Re: Number theory #1

¡Ay chico, estás hablando Chino!

So, is (mod a+b) a modulus operation? And why bother if you multiply it by zero? And how can you define the product of two natural numbers to be zero?

What's a rom (besides a read-only memory)?

Or were you joking?

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Number theory #1

I've never seen rom either.  Could we get a definition, krassi?

So, is (mod a+b) a modulus operation? And why bother if you multiply it by zero? And how can you define the product of two natural numbers to be zero?

modulo, to be exact.  But it is the same idea.

k ≡ m (mod n) means that n divides k-m:

k-m = xn, x ∈ Z

So in this situation, a+b divides ab.  Since a+b divides ab, a+b divides (ab)² or a²b².

Not knowing rom, I can't go any further than that.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: Number theory #1

Rom is the capital of Ital

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Number theory #1

*slaps MathIsFun*

Everyone knows that Rom is the capital of rom.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

darthradius

Member

Registered: 2005-11-28

Posts: 97

Re: Number theory #1

hey krassi-
I'm still waiting to hear what rom is...

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
                                                             -Bertrand Russell

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Number theory #1

Oh, a stupid muistake! not "rom"- it's "form". It means that

Last edited by krassi_holmz (2006-02-11 03:47:00)

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Number theory #1

for some k∈Z

Now let a + b = y^3 and k = x.  Then:

Plugging this back in...:

So....

Which was to be shown.

Last edited by Ricky (2006-02-11 05:48:32)

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Number theory #1

"Now let a + b = y^3 and k = x"-
you cannot let a+b=y^3, because then y may not be integer.l

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Number theory #1

Yea... I was hoping you wouldn't catch that.

Do you know the solution?

Last edited by Ricky (2006-02-11 09:57:33)

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Number theory #1

I know it but i want you to get it.

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darthradius

Member

Registered: 2005-11-28

Posts: 97

Re: Number theory #1

Well, krassi...I think its safe to say that none of us came up with an answer and I've been really wanting to see the solution...
So can you post it, please?

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
                                                             -Bertrand Russell

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Number theory #1

Give me till tuesday, I think I'm close, but I have a test in Advanced Calc and Modern Algebra that I got to get through first on monday.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Number theory #1

OK. I'll wait for some time.
The proof is not so hard.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Number theory #1

Alright, I'm bout ready to give up.  It seems like I'm missing some fact that being 0 mod gives, but I can't seem to find it.  Being divisible by a+b doesn't seem to be enough.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

darthradius

Member

Registered: 2005-11-28

Posts: 97

Re: Number theory #1

In that case....give us the proof please, krassi!!!!
It's been bugging me since you posted this and I'm having the same problem...I've gone through all my notes to try to find some long lost number theory theorem that would make it all come together, but I can't....and I'm a bit obsessive so I need to see the proof...and I need to see it fast!

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
                                                             -Bertrand Russell

damathamatician

Member

Registered: 2006-02-28

Posts: 10

Re: Number theory #1

what do the buttons do?

Last edited by damathamatician (2006-02-28 10:58:15)

krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Number theory #1

Let
d=GCD[a,b]
Then
a=di and b=dj and GCD[i,j]=1.
ab≡0(mod a+b) =>
(di)(dj)≡0(mod d(i+j))
dij≡0(mod (i+j)) ->>(i)
But GCD[i,j]=1
GCD[i,i+j]=1
GCD[j,i+j]=1
=>
GCD[ij,i+j]=1
but from (i) (i+j)|dij => (i+j)|d => exists k ∈ N : k(i+j)=d

Let go back to the main equation:

where x=k; y=ij(i+j)

Last edited by krassi_holmz (2006-02-28 19:43:21)

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krassi_holmz

Real Member

Registered: 2005-12-02

Posts: 1,905

Re: Number theory #1

NT#2:
Prove that for every a,b

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