Obedient figures/numbers

[enoimreh7]

Lisa calls a natural number obedient if she can be written as a product of the two figures following on each other. Example: The number 20 is obedient, because 20 = 4∙5.
Points: To every obedient number there is at least one second obedient number which proves an obedient number with the first one multiplied again.
Example: 2∙3=6 and 3∙4=12 and 6∙12=72 which is: 8∙9
Is the sentence true??Why?? Is there a proof?

 

[enoimreh7]

Obedient numbers

Lara calls a natural number obedient if she can be written as a product of the two figures following on each other. Example: The number 20 is obedient, because 20 = 4∙5. Points: To every obedient number there is at least one second obedient number which proves an obedient number with the first ones multiplied again. 2 times 3 is 6 // 3 times 4 is twelve // 6 times twelve is 72 and 72 is 8 times 9 // Is there an Abstract proof for this sentence?

 

[Dr.Peterson]

I'm not at all sure I understand the question.

As I understand it, an "obedient number" is one that is the product of two consecutive integers, n(n+1) for some integer n.

But how is the second number supposed to be related to the first? What does "with the first one multiplied again" mean?

Possibly you mean that if m = n(n+1) is an obedient number, then m times something else will be an obedient number as well; but that is obviously true, since m(m+1) will be obedient. Your example looks like you must mean something more than that.

 

[tkhunny]

"two figures following each other"

First, this is impossible without a closed path (maybe a circle) -- Better definition, please.

Second, (Dog Stick Figure) => (Cat Stick Figure) -- (running in a circle)

Demonstrate this multiplication.

Seriously improved definition and problem statement, please.

 

[enoimreh7]

Obedient numbers

I'm not at all sure I understand the question.

As I understand it, an "obedient number" is one that is the product of two consecutive integers, n(n+1) for some integer n.

But how is the second number supposed to be related to the first? What does "with the first one multiplied again" mean?

Possibly you mean that if m = n(n+1) is an obedient number, then m times something else will be an obedient number as well; but that is obviously true, since m(m+1) will be obedient. Your example looks like you must mean something more than that.

Hello Dr. Peterson!
Thank you for your answer! The exercise wants me to prove the following:
Each obedient number m=n(n+1) can be multiplied with at least one other obedient number like o=p(p+1) or in different words for each m exists at least one other obedient number like for example o which than, multiplied with the first obedient number, will produce a third obedient number. So m times o is y and y=x(x+1)
How the second obedient number is related to the first one is not really defined, only with: there is at least one second obedient number which will - multiplied with the first obedient number - produce a third obedient number.
I just found an example that matches with 2 times 3 is 6 and 3 times 4 is 12 Those two obedient numbers: 6 times 12 produce 72 which is a third obedient number whereas 8 times 9 is 72.
I hope this shows the Problem in a better way!?
Sincerly, Yours Enoimreh

 

[enoimreh7]

dog and cat

"two figures following each other"

First, this is impossible without a closed path (maybe a circle) -- Better definition, please.

Second, (Dog Stick Figure) => (Cat Stick Figure) -- (running in a circle)

Demonstrate this multiplication.

Seriously improved definition and problem statement, please.

They both know karate, will stop to run and decide that they do not need a circle, because of their respect for each other.

 

[tkhunny]

Nice try, but they are still different species and multiplication is unlikely.

So, m(m+1) * n(n+1) = q(q+1)

Well, there are only 13 choices for q. Check them all out. Can the LHS be so factored?

 

[enoimreh7]

Nice try, but they are still different species and multiplication is unlikely.

So, m(m+1) * n(n+1) = q(q+1)

Well, there are only 13 choices for q. Check them all out. Can the LHS be so factored?

Thank you for trying. I followed your idea till I found the twenteeth possibility.

m(m+1) * n(n+1) = x = q(q+1)

01*02=02 02*03=06 12=03*04
02*03=06 03*04=12 72=08*09
03*04=12 04*05=20 240=15*16
20xxxxxxx 05*06=30 600=24*25
30xxxxxxx 06*07=42 1260=35*36
42xxxxxxx 07*08=56 2352=48*49
56xxxxxxx 08*09=72 4032=63*64
72xxxxxxx 09*10=90 6480=80*81
90xxxxxxx 10*11=110 9900=99*100
110 * 132= 14520=120*121
132 * 156= 20592=143*144
156 * 182= 28392=168*169
182 * 210= 38220=195*196
210 * 240= 50400=224*225
240 * 272= 65280=255*256
272 * 306= 83232=288*289
306 * 342=104652=323*324
342 * 380=129960=360*361
380 * 420=159600=399*400
420 * 462=194040=440*441

What I can see is, that if
m(m+1) * n(n+1) = x = q(q+1)
than
m(m+1) is related to q because q is:
first m(m+1) plus 1
second m(m+1) plus 2
third m(m+1) plus 3
and so forth.
twenteeth possibility I showed q is m(m+1) plus 20

I think that continues ..... and is infinite
but how do I put this in a proof????????

And also this is only one way to find the second obedient number, it could also be found in a different way like I don't know how, but I have to exclude it in a proof.

Could you please help me to find the equotation with one or even two unknown ????
Thank you very much, Sincerly, E.

And what does LHS mean?

 

[tkhunny]

And what does LHS mean?

LHS = Left Hand Side

Not a proof. One must show them ALL.

 

[enoimreh7]

LHS = Left Hand Side

Not a proof. One must show them ALL.

First an equation, than use the equation. Do you really mean there is no equation? Sincerly, E.

 

[enoimreh7]

Hello Dr. Peterson!
Thank you for your answer! The exercise wants me to prove the following:
Each obedient number m=n(n+1) can be multiplied with at least one other obedient number like o=p(p+1) or in different words for each m exists at least one other obedient number like for example o which than, multiplied with the first obedient number, will produce a third obedient number. So m times o is y and y=x(x+1)
How the second obedient number is related to the first one is not really defined, only with: there is at least one second obedient number which will - multiplied with the first obedient number - produce a third obedient number.
I just found an example that matches with 2 times 3 is 6 and 3 times 4 is 12 Those two obedient numbers: 6 times 12 produce 72 which is a third obedient number whereas 8 times 9 is 72.
I hope this shows the Problem in a better way!?
Sincerly, Yours Enoimreh

Dear Sir!
When you answered to a question - do you than follow the conversation - or are you just reading the answers?
Here is the next ,,working and thinking step" and now I'm stuck. Please have a look:

m(m+1) * n(n+1) = x = q(q+1)

01*02=02 02*03=06 12=03*04
02*03=06 03*04=12 72=08*09
03*04=12 04*05=20 240=15*16
20xxxxxxx 05*06=30 600=24*25
30xxxxxxx 06*07=42 1260=35*36
42xxxxxxx 07*08=56 2352=48*49
56xxxxxxx 08*09=72 4032=63*64
72xxxxxxx 09*10=90 6480=80*81
90xxxxxxx 10*11=110 9900=99*100
110 * 132= 14520=120*121
132 * 156= 20592=143*144
156 * 182= 28392=168*169
182 * 210= 38220=195*196
210 * 240= 50400=224*225
240 * 272= 65280=255*256
272 * 306= 83232=288*289
306 * 342=104652=323*324
342 * 380=129960=360*361
380 * 420=159600=399*400
420 * 462=194040=440*441

What I can see is, that if
m(m+1) * n(n+1) = x = q(q+1)
than
m(m+1) is related to q because q is:
first m(m+1) plus 1
second m(m+1) plus 2
third m(m+1) plus 3
and so forth.
twenteeth possibility I showed q is m(m+1) plus 20

I think that continues ..... and is infinite
but how do I put this in a proof????????

And also this is only one way to find the second obedient number, it could also be found in a different way like I don't know how, but I have to exclude it in a proof.

Could you please help me to find the equation with one or even two unknown ????
Sincerly, Yours E. (see / hear you tomorrow - good bye)

 

[Dr.Peterson]

I've been checking in when I am free, to see what others have said, and whether there is anything new to help me understand the problem. I find it very difficult to guess what you are thinking in that big block of numbers.

But I think what you are saying is that you have found by experiment that the product of two CONSECUTIVE "obedient numbers" always seems to be an obedient number; that is, if you are given m(m+1), you can take the other one to be (m+1)(m+2), where you have increased each of the factors by 1, and their product appears to be an obedient number. Is that what you are saying?

If so, then what you need to do to make an algebraic proof is to express the product, m(m+1)*(m+1)(m+2), as a product of two consecutive integers. Can you do that?

 

[JeffM]

I too am unsure I understand the problem, particularly given the forum it is posted in. If this is a problem given to students who do not know algebra, the answer will not make sense to the student. It does not seem to be a problem in arithmetic. (In the 19th century, number theory was called Higher Arithmetic.)

In any case, if the problem is to show that the product of two obedient numbers is an obedient number, where an obedient number is the product of an integer and its successor, then there are obviously an infinite number of obedient numbers.

Is this the problem:

$\displaystyle \text {GIVEN }a = x(x + 1) = x^2 + x \text {, and } b = y^2 + y \text {, where } x,\ y \in \mathbb Z \text {, PROVE}$

$\displaystyle \exists\ z \in \mathbb Z \text {, such that } z(z + 1) = c = ab.$

EDIT: Obviously 2 is an obedient number because 1 * (1 + 1) = 2. However, 2 * 2 is not an obedient number so the proposition given above is false. However, 2 and 6 are obedient numbers, and their product is an obedient number so clearly some products of obedient numbers are obedient. This takes us right back to what in Heaven's name is the statement of the problem?

 

[Dr.Peterson]

As I understand it, the problem is:

$\displaystyle \text {GIVEN }a = x(x + 1) \text {, where } x \in \mathbb Z \text {, PROVE }\exists\ y,\ z \in \mathbb Z \text {, }$
$\displaystyle \text{such that } x(x + 1) \cdot y(y + 1) = z(z + 1).$

That is, the challenge is to prove that for ANY obedient number a, there is SOME other obedient number b such that ab is also an obedient number. The conjecture (which is true and easy to prove) is that this can be refined by taking b as the next consecutive obedient number; that is,

$\displaystyle \text {GIVEN }a = x(x + 1) \text {, where } x \in \mathbb Z, \text { and } b = (x + 1)(x + 2) \text{, THEN } \exists\ z \in \mathbb Z \text{,}$
$\displaystyle \text {such that } ab = z(z + 1).$

 

[enoimreh7]

As I understand it, the problem is:

$\displaystyle \text {GIVEN }a = x(x + 1) \text {, where } x \in \mathbb Z \text {, PROVE }\exists\ y,\ z \in \mathbb Z \text {, }$
$\displaystyle \text{such that } x(x + 1) \cdot y(y + 1) = z(z + 1).$

That is, the challenge is to prove that for ANY obedient number a, there is SOME other obedient number b such that ab is also an obedient number. The conjecture (which is true and easy to prove) is that this can be refined by taking b as the next consecutive obedient number; that is,

$\displaystyle \text {GIVEN }a = x(x + 1) \text {, where } x \in \mathbb Z, \text { and } b = (x + 1)(x + 2) \text{, THEN } \exists\ z \in \mathbb Z \text{,}$
$\displaystyle \text {such that } ab = z(z + 1).$

Thank you very much Dr. Peterson!
And z always is x(x+1) + x so that ab = (x (x+1) +x) * ((x (x+1) +x)+1)

Is that correct? And is this a proof?

I recognized this, looking at my big block of numbers which is showing:
x(x+1)=a than y(y+1)=b than product of ab=z(z+1)

Could than the answer to this not really clear question be:
Each obedient number has at least one other obedient number with which it will when multiplied produce a third obedient number. This is infinite.

 

[enoimreh7]

I too am unsure I understand the problem, particularly given the forum it is posted in. If this is a problem given to students who do not know algebra, the answer will not make sense to the student. It does not seem to be a problem in arithmetic. (In the 19th century, number theory was called Higher Arithmetic.)

In any case, if the problem is to show that the product of two obedient numbers is an obedient number, where an obedient number is the product of an integer and its successor, then there are obviously an infinite number of obedient numbers.

Is this the problem:

$\displaystyle \text {GIVEN }a = x(x + 1) = x^2 + x \text {, and } b = y^2 + y \text {, where } x,\ y \in \mathbb Z \text {, PROVE}$

$\displaystyle \exists\ z \in \mathbb Z \text {, such that } z(z + 1) = c = ab.$

EDIT: Obviously 2 is an obedient number because 1 * (1 + 1) = 2. However, 2 * 2 is not an obedient number so the proposition given above is false. However, 2 and 6 are obedient numbers, and their product is an obedient number so clearly some products of obedient numbers are obedient. This takes us right back to what in Heaven's name is the statement of the problem?

Thank you JeffM! I didn't recognize 2*2 (?) I continued, pleas look at my answer to Dr.P. Thank you, Sincerly,E.

 

[Dr.Peterson]

Thank you very much Dr. Peterson!
And z always is x(x+1) + x so that ab = (x (x+1) +x) * ((x (x+1) +x)+1)

Is that correct? And is this a proof?

I recognized this, looking at my big block of numbers which is showing:
x(x+1)=a than y(y+1)=b than product of ab=z(z+1)

Could than the answer to this not really clear question be:
Each obedient number has at least one other obedient number with which it will when multiplied produce a third obedient number. This is infinite.

First, what I did was to restate what I think the problem was, and what your claim was. Are you confirming this?

What you have written is not really a proof of your claim, just a new claim apparently based on a few examples, which as you know does not constitute a proof. Can you convince me that it is true?

The simple proof I had in mind was just to take ab = [x(x+1)][(x+1)(x+2)] and rewrite it in an appropriate form. If your claim is valid, then you should be able to rearrange it as [x(x+1)+x][(x(x+1)+x)+1]; but my way of doing it is simpler. I was helped by knowledge of a pattern involving groups of consecutive numbers, namely that, for example, 2*4 = (3-1)(3+1) = 3^2 - 1. This is easily proved by algebra, and is in fact equivalent to what you need to prove here.

Your final statement here, "This is infinite", is probably not what you mean; no number can be infinite. Do you mean to say that there are infinitely many such numbers? That adds nothing to the problem, as it is already implied by the claim that it is true for ALL obedient numbers (of which there are infinitely many). I think all you are doing in your last paragraph is restating the problem, not giving an answer.

 

[Dr.Peterson]

Thank you JeffM! I didn't recognize 2*2 (?) I continued, pleas look at my answer to Dr.P. Thank you, Sincerly,E.

Note that JeffM's interpretation of the problem was that ANY product of two obedient numbers is obedient, and his example was a counterexample for that claim. It is not relevant to your actual problem, if I understand it correctly.

 

[JeffM]

I am guessing that you are not a native speaker of English and that you perhaps are not conversant with some aspects of mathematical notation that facilitate talking across different languages.

$\displaystyle a \in \mathbb Z \text { and } b = a(a + 1) \iff b \text { is an obedient number.}$

It is provable that the number of obedient numbers is infinite.

Dr. Peterson seems to have deciphered your problem. Yes, given any obedient number p, there is at least one other obedient number q such that pq is an obedient number. And he has given you the basis of a proof.

$\displaystyle p \text { is an obedient number} \implies \exists \ x \in \mathbb Z \text { such that } x(x + 1) = p.$

$\displaystyle \text {Let } y = x + 1 \implies x = y - 1 \implies p =(y - 1)y.$

$\displaystyle \text {Let } q = y(y + 1) \implies q \text { is an obedient number.}$

$\displaystyle \therefore pq = (y - 1)y * y(y + 1) = y^2(y^2 - 1) = (y^2 - 1)\{(y^2 - 1) + 1\} \implies$

$\displaystyle pq \text { is an obedient number.}$

Very elegant Dr. P.

Does this answer your question?