Are These Rules new Conjectures in Set Theory?

[Gh. Soleimani]

We can easily find out below rules in set theory:

1. Let consider set “A” as follows:

A = {a1, a2, a3, a4… an} and also power set of A is set C:

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

Rule 1: To find the number of subsets with precise members number, we can use Binomial Coefficient

C (n, r) = n! / r! (n-r)!

Where:

n = number of members in set A

r = number of precise member for each subset

Example:

We have set A = {1, 2, 3, 4, 5, 6, 7}

The number of subsets with no member is: C (7, 0) = 1

The number of subsets with one member is: C (7, 1) = 7

The number of subsets with two member is: C (7, 2) = 21

The number of subsets with three member is: C (7, 3) = 35

The number of subsets with four member is: C (7, 4) = 35

The number of subsets with five member is: C (7, 5) = 21

The number of subsets with six member is: C (7, 3) = 7

The number of subsets with seven member is: C (7, 3) = 1

2. Let consider set “A” and power set of A which is set “C” as follows:

A = {a1, a2, a3, a4… an}

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” below cited:

B = {x, y, z, t,…..}

Where:

x, y, z, t,… = functions of internal sum of each subset

For instance, x = a1, z = a1 + a2, t = a1 + a2 + a3 and ……..

Rule 2: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

SB / SA = 2^ (n-1) , n = number of members set A

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

SB = 0+1+2+3+4+3+4+5+5+6+7+6+7+8+9+10 = 80

SA = 1+2+3+4 = 10

SB / SA = 2^ (n-1) = 2 ^ (4-1) = 80 /10

Now, the questions is:

- Are these rules new ones in Set Theory?

- Are These Rules new Conjectures in Mathematics?

 

[andrewkirk]

What you are doing looks more like Combinatorics that Set Theory.

We can find set “B” below cited:

B = {x, y, z, t,…..}

Where:

x, y, z, t,… = functions of internal sum of each subset

For instance, x = a1, z = a1 + a2, t = a1 + a2 + a3 and ……..

This needs to be explained better, because it's not at all clear what you intend B to be. You can't define a set with a 'for instance'. You need to define it using recognised, unambiguous set operations such as powerset, union, intersection, complementation or filtering on a logical criterion. Enumeration (listing elements) is fine, as long as it's completely clear what the elements are, but it's not OK if you use ellipses (the '...' symbol) as you have above. Enumeration has to be complete, with no ellipses and no 'etc'.

 

[Gh. Soleimani]

What you are doing looks more like Combinatorics that Set Theory.

This needs to be explained better, because it's not at all clear what you intend B to be. You can't define a set with a 'for instance'. You need to define it using recognised, unambiguous set operations such as powerset, union, intersection, complementation or filtering on a logical criterion. Enumeration (listing elements) is fine, as long as it's completely clear what the elements are, but it's not OK if you use ellipses (the '...' symbol) as you have above. Enumeration has to be complete, with no ellipses and no 'etc'.

Sorry, it is not the answer to my questions.

 

[Samy_A]

No.

Rule 1 is well known (https://en.wikipedia.org/wiki/Binomial_coefficient#Combinatorics_and_statistics)
Rule 2 (if I interpreted it correctly, @andrewkirk correctly stated that you are not very clear there), is a consequence of the well known result that the power set of a set with $n$ elements has $2^n$ elements (https://en.wikipedia.org/wiki/Power_set#Properties).

 

[Gh. Soleimani]

No.

Rule 1 is well known (https://en.wikipedia.org/wiki/Binomial_coefficient#Combinatorics_and_statistics)
Rule 2 (if I interpreted it correctly, @andrewkirk correctly stated that you are not very clear there), is a consequence of the well known result that the power set of a set with $n$ elements has $2^n$ elements (https://en.wikipedia.org/wiki/Power_set#Properties).

Please care again on rule1. Your answer is completely wrong. I say to you a problem: How can we find the number of subsets with precise members number? The solution is, to use Binomial Coefficient. For instance, if we have a set with 7 members, The number of subsets with four member will be 35. Have you any anouther solution for this problem? or Have you another reference which shows us this solution for this problem?

 

[Mark44]

Please care again on rule1. Your answer is completely wrong.

Samy_A was responding to what he thought you meant. What you stated was not clear, especially the part "with precise members number":

Rule 1: To find the number of subsets with precise members number, we can use Binomial Coefficient

The binomial coefficients give the sizes of each of the subsets of a certain size.

I say to you a problem: How can we find the number of subsets with precise members number? The solution is, to use Binomial Coefficient.

Which is what Samy_A was saying.

For instance, if we have a set with 7 members, The number of subsets with four member will be 35. Have you any anouther solution for this problem? or Have you another reference which shows us this solution for this problem?

 

[Gh. Soleimani]

Samy_A was responding to what he thought you meant. What you stated was not clear, especially the part "with precise members number":

The binomial coefficients give the sizes of each of the subsets of a certain size.
Which is what Samy_A was saying.

Thank you for your reply. But I did not see your quote (The binomial coefficients give the sizes of each of the subsets of a certain size) on this link: https://en.wikipedia.org/wiki/Binomial_coefficient#Combinatorics_and_statistics whereas, I need to have official reference (academic reference) because I am willing to know if your quote had been already proved or it is a conjecture. In the meanwhile, please be informed that wikipedia is not a reference because each moment its content has been edited.

 

[Mark44]

Thank you for your reply. But I did not see your quote (The binomial coefficients give the sizes of each of the subsets of a certain size) on this link: https://en.wikipedia.org/wiki/Binomial_coefficient#Combinatorics_and_statistics

It's in that section that Samy_A cited. For your example, with a set with 7 elements, the number of subsets with 4 elements is $\binom{7}{4}$ = 35.

If you have a set with n elements, the number of subsets with exactly k elements is $\binom{n}{k} = \frac{n!}{k!(n - k)!}$.

whereas, I need to have official reference (academic reference) because I am willing to know if your quote had been already proved or it is a conjecture. In the meanwhile, please be informed that wikipedia is not a reference because each moment its content has been edited.

These formulas for the binomial coefficients have been around for a long time, and are very well known. Any textbook on combinatorics or probability will have this formula.

 

[Gh. Soleimani]

It's in that section that Samy_A cited. For your example, with a set with 7 elements, the number of subsets with 4 elements is $\binom{7}{4}$ = 35.

If you have a set with n elements, the number of subsets with exactly k elements is $\binom{n}{k} = \frac{n!}{k!(n - k)!}$.

These formulas for the binomial coefficients have been around for a long time, and are very well known. Any textbook on combinatorics or probability will have this formula.

Yes. Thank you. I found the proof of rule 1 on below papers:
http://www.maths.qmul.ac.uk/~pjc/notes/comb.pdf
http://math.mit.edu/~rstan/ec/ec1.pdf

 

[Gh. Soleimani]

Let me tell you that the basic of above rules returns back to my case of accounting control such as below example:
Consider: Set A = {a1, a2, a3, a4,...} and set B = {b1, b2, b3, b4,...]
Now,we want to know: which total sum of 1 or 2 or 3 or more of members of set A are just equal to total sum of 1 or 2 or 3 or more of members of set B? For instance,
a1 = b2+b3+b4+b5 or a2+a3+a7+a12 = b2+b3 and etc. How can we find all equilibriums?
I solved this case by using of a idea of power set and posted my solution on my website. Do you have any another idea to solve this case?

 

[Mark44]

Rule 2: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:
SB / SA = 2^ (n-1) , n = number of members set A

Your rule doesn't work.
Counterexample :
Let A = {1, 2}
P(A) = { {}, 1, 2, {1, 2}}
Let B = { 2, 3, 5, 6}
P(B) = { {}, 2, 3, 5, 6, {2, 3}, {2, 5}, {2, 6}, {3, 5}, {3, 6}, {5, 6}, {2, 3, 5}, {2, 3, 6}, {2, 5, 6}, {3, 5, 6}, {2, 3, 5, 6}}
$\sum{P(B}$ =128

$\sum{P(A}$ = 1 + 2 + 3 = 6
$\frac{\sum{P(B}}{\sum{P(A}} = \frac{128}{6} \approx 21.33$. This is not a power of 2 for any integer n.

 

[Samy_A]

It's not very clear what rule 2 exactly means.

My interpretation was as follows:
A is a finite set of integers with n elements. SA is the sum of all elements of A.
C is the power set of A.
Now, for each element of C one can compute the sum of its elements: these sums are the element of the set (actually the multiset) B.
SB is the sum of all elements of B.

Rule 2 states than SB=SA*$2^{(n-1)}$
I think that with this interpretation the rule is correct, but quite elementary.
As every element of A will be an element of exactly $2^{(n-1)}$ subsets of A, it is rather obvious that SB=SA*$2^{(n-1)}$

 

[Gh. Soleimani]

Mark 44: Your way to get elements of B in your counterexmample is wrong. Because, we will have 0, 1, 2 and 3 for B then SB = 6 and SA = 3 , SB/SA = 2 = 2 ^ 2 -1
Samy A: you got true concept of this rule. Do you have any proof for this rule or it is a conjecture?

 

[Mark44]

Mark 44: Your way to get elements of B in your counterexmample is wrong.

No, it is not wrong. You have not specified that the sets have to be consecutive integers starting with 1.
Rule 2 just mentions two sets A and B, and doesn't put any further restrictions on them.
It's up to you to provide more conditions.

Because, we will have 0, 1, 2 and 3 for B then SB = 6 and SA = 3 , SB/SA = 2 = 2 ^ 2 -1
Samy A: you got true concept of this rule. Do you have any proof for this rule or it is a conjecture?

 

[Samy_A]

Mark 44: Your way to get elements of B in your counterexmample is wrong. Because, we will have 0, 1, 2 and 3 for B then SB = 6 and SA = 3 , SB/SA = 2 = 2 ^ 2 -1

It is up to you to be clear. I was "lucky" in guessing what you meant with rule 2, mainly because I worked backward from the result.

Samy A: you got true concept of this rule. Do you have any proof for this rule or it is a conjecture?

I already stated the "proof":

As every element of A will be an element of exactly $2^{(n-1)}$ subsets of A, it is rather obvious that SB=SA*$2^{(n-1)}$

This is just a consequence of the well known result that the power set of a set with m elements has $2^{m}$ elements, applied to m=n-1. (As already stated in my first post in this thread, by the way.)

 

[Gh. Soleimani]

I deleted rule 1 and edited rule 2 and also added new rule (rule 3) as follows:

1. Let consider set “A” and power set of A which is set “C” as follows:

A = {a1, a2, a3, a4… an}

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by adding members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {a1+ a2}, {a1 + a3}, {a2 + a3 + a4)….{an}}Rule 1: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

SB / SA = 2^ (n-1) , n = number of members set A

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {1+2}, {1+3}, {1+4}, {2+3}, {2+4}, {3+4}, {1+2+3}, {1+2+4}, {1+3+4}, {2+3+4}, {1+2+3+4}}

SB = 0+1+2+3+4+3+4+5+5+6+7+6+7+8+9+10 = 80

SA = 1+2+3+4 = 10

SB / SA = 2^ (n-1) = 2 ^ (4-1) = 80 /10 2. Let consider set “A” and power set of A which is set “C” as follows:

A = {a1, a2, a3, a4… an}

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by multiplying members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {a1* a2}, {a1*a3}, {a2*a3*a4)….{an}}

Rule 2: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

(SB+ 1)/ SA = 2n!/n , n = number of members set A

Or SB = [(2n!/n)*SA] - 1

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {1*2}, {1*3}, {1*4}, {2*3}, {2*4}, {3*4}, {1*2*3}, {1*2*4}, {1*3*4}, {2*3*4}, {1*2*3*4}}

SA = 1+2+3+4 = 10 , SB = 1+2+3+4+2+3+4+6+8+12+6+8+12+24+24 =119

(SB +1) / SA = (2*(1*2*3*4)/4) = 12 = (119 +1)/10 Now, the questions is:

- Are these rules new ones in Set Theory?

- Are These Rules new Conjectures in Mathematics?

 

[Samy_A]

Now, the questions is:

- Are these rules new ones in Set Theory?

- Are These Rules new Conjectures in Mathematics?

You have already been shown that Rule 1 (your former Rule 2) is a trivial consequence of an elementary fact about power sets.

Rule 2 is wrong without further conditions on A.
Example:
n=2
A={2,5}
SA=7
C={{},{2},{5},{2,5}}
B={0,2,5,10}
SB=17
(SB+ 1)/ SA =18/7
2n!/2=2

 

[Gh. Soleimani]

Yes. You are true.
A = a set of Natural number
Of Course, I do not consider zero as Natural number. I will say it Whole number if we consider zero.

 

[Samy_A]

Yes. You are true.
A = a set of Natural number
Of Course, I do not consider zero as Natural number. I will say it Whole number if we consider zero.

My counterexample A={2,5} is a set of natural numbers, and doesn't contain 0. So your new rule 2 is still wrong.

 

[Gh. Soleimani]

I did not tell you that A is a subset of natural number. It is ok. Let us consider N instead of A. It means that set of natural number. and then we have:
(SB+1) / SN = 2n!/n

 

[Samy_A]

I did not tell you that A is a subset of natural number. It is ok. Let us consider N instead of A. It means that set of natural number. and then we have:
(SB+1) / SN = 2n!/n

Can you please state clearly the conditions that the set A (or whatever you want to call it, the name of the set is irrelevant) has to satisfy in order for your rule 2 to be valid?

Clearly rule 2 is not valid for the set {2,5}.

If you don't state all the conditions for your rule 2 clearly, it is meaningless.

 

[Gh. Soleimani]

please wait. I am busy now

 

[Mark44]

Can you please state clearly the conditions that the set A (or whatever you want to call it, the name of the set is irrelevant) has to satisfy in order for your rule 2 to be valid?

Clearly rule 2 is not valid for the set {2,5}.

If you don't state all the conditions for your rule 2 clearly, it is meaningless.

I agree completely. If we decide to try your (Gh. Soleimani) rule on, say A = {1, 3, 5} and B = {2}, and find that it doesn't work, you can't come back and tell us we are wrong. As Samy_A said, if you don't specify clearly any and all conditions on the sets involved, the rules are meaningless.

 

[Gh. Soleimani]

I edited my article and my assumtions as follows:

2. Let consider A as set of Arithmetic Progression where:

d = 1, a1 = 1 and an = a1 + (n - 1)d, n = 1, 2, 3,…..

In this case, we have:

A = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1)d)}

And power set of A which is set “C” as follows:

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by multiplying members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {a1* a2}, {a1*a3}, {a2*a3*a4)….{an}}

Rule 2: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

(SB+ 1)/ SA = 2n!/n , n = number of members set A

Or SB = [(2n!/n)*SA] - 1

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {1*2}, {1*3}, {1*4}, {2*3}, {2*4}, {3*4}, {1*2*3}, {1*2*4}, {1*3*4}, {2*3*4}, {1*2*3*4}}

SA = 1+2+3+4 = 10 , SB = 1+2+3+4+2+3+4+6+8+12+6+8+12+24+24 =119

(SB +1) / SA = (2*(1*2*3*4)/4) = 12 = (119 +1)/10Now, the questions is:

- Are these rules new ones in Set Theory?

- Are These Rules new Conjectures in Mathematics?

 

[Samy_A]

I edited my article and my assumtions as follows:

2. Let consider A as set of Arithmetic Progression where:

d = 1, a1 = 1 and an = a1 + (n - 1)d, n = 1, 2, 3,…..

In this case, we have:

A = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1)d)}

And power set of A which is set “C” as follows:

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by multiplying members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {a1* a2}, {a1*a3}, {a2*a3*a4)….{an}}

Rule 2: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

(SB+ 1)/ SA = 2n!/n , n = number of members set A

Or SB = [(2n!/n)*SA] - 1

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {1*2}, {1*3}, {1*4}, {2*3}, {2*4}, {3*4}, {1*2*3}, {1*2*4}, {1*3*4}, {2*3*4}, {1*2*3*4}}

SA = 1+2+3+4 = 10 , SB = 1+2+3+4+2+3+4+6+8+12+6+8+12+24+24 =119

(SB +1) / SA = (2*(1*2*3*4)/4) = 12 = (119 +1)/10Now, the questions is:

- Are these rules new ones in Set Theory?

- Are These Rules new Conjectures in Mathematics?

Thanks.

Let's simplify this for starters.
For a given n, A is simply the set of all integers from 1 to n: A={1,2,...,n}
So, SA=n(n+1)/2

Your rule then simplifies to $SB=\frac{2n!}{n}*\frac{n(n+1)}{2} - 1=n!(n+1)-1=(n+1)!-1$

So, for a given n, the rule states: SB=(n+1)!-1

I'm not immediately aware of this rule being known (someone else may be, of course), but it can be proven easily. I did it by mathematical induction, by the way.

 

[Gh. Soleimani]

Thank you to simply this rule. Then you say that this rule is not a conjecture.

 

[Samy_A]

Thank you to simply this rule. Then you say that this rule is not a conjecture.

A conjecture is a mathematical statement for which no proof is known. Fermat's Last Theorem was a conjecture for centuries, but now is a (proven) theorem.
Your rule is not a conjecture, as it can be proven.
Whether it is new or not, whether it is useful or not, I can't really say.

 

[Gh. Soleimani]

I think the last conjecture is referred to Beal Conjecture and I remember AMS awarded $1000000 prize for someone who presented a proof or counterexample.
Anyway, here is another new rule:3. Let consider set “A” as subset of R and power set of A which is set “C” as follows:
A ⊂ R
C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by average of members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {(a1+ a2)/2}, {(a1 + a3)/2}, {(a2 + a3 + a4)/3}….{an}}Rule 3: If SB = total Sum of members of set B:

Then: SB = ((2^n)-1)*Average (A) , n = number of members set A

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {(1+2)/2}, {(1+3)/2}, {(1+4)/2}, {(2+3)/2}, {(2+4)/2}, {(3+4)/2}, {(1+2+3)/3}, {(1+2+4)3}, {(1+3+4)/3}, {(2+3+4)/3}, {(1+2+3+4)/4}

SB = 0+1+2+3+4+1.5+2+2.5+2.5+3+3.5+2+2.333333+2.6666666+3+2.5 = 37.5

Average (A) = (1+2+3+4)/4 = 2.5

SB = ((2^ 4) -1)*2.5 = 37.5

 

[Samy_A]

I think the last conjecture is referred to Beal Conjecture and I remember AMS awarded $1000000 prize for someone who presented a proof or counterexample.

I don't understand what all this has to do with Beal's conjecture.

Anyway, here is another new rule:3. Let consider set “A” as subset of R and power set of A which is set “C” as follows:
A ⊂ R
C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by average of members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {(a1+ a2)/2}, {(a1 + a3)/2}, {(a2 + a3 + a4)/3}….{an}}Rule 3: If SB = total Sum of members of set B:

Then: SB = ((2^n)-1)*Average (A) , n = number of members set A

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {(1+2)/2}, {(1+3)/2}, {(1+4)/2}, {(2+3)/2}, {(2+4)/2}, {(3+4)/2}, {(1+2+3)/3}, {(1+2+4)3}, {(1+3+4)/3}, {(2+3+4)/3}, {(1+2+3+4)/4}

SB = 0+1+2+3+4+1.5+2+2.5+2.5+3+3.5+2+2.333333+2.6666666+3+2.5 = 37.5

Average (A) = (1+2+3+4)/4 = 2.5

SB = ((2^ 4) -1)*2.5 = 37.5

Nice one.

The proof is straightforward using the well known identity $\sum_{k=0}^n \binom{n}{k}=2^n$.

 

[Gh. Soleimani]

The Beal Conjecture states that the only solutions to the equation A^x + B^y = C^z, when A, B and C are positive integers, and x, y and z are positive integers greater than two, are those in which A, B and C have a common factor. The American Mathematical Society in Providence, Rhode Island, said that typical of many statements in number theory, they're "easy to say but extremely difficult to prove.” The Beal Prize is funded by D. Andrew Beal, a prominent banker who is also a mathematics enthusiast. An AMS-appointed committee, the Beal Prize Committee, will recommend awarding this prize for either a proof or a counterexample of the Beal Conjecture published in a refereed and respected mathematics publication. The prize money—now US$1,000,000—is being held by the AMS until it is awarded. The spendable income from investment of the prize money is used to fund the annual Erdős Memorial Lecture and other activities of the Society that benefit early-career mathematicians.

A valid counterexample needs to satisfy all four conditions—don't leave one out:

A,B,C,x,y,z are positive integers
x,y,z>2
A^x+B^y=C^z
A,B,C have no common prime factor.