A simple equation with a complicated property
Let, $Bbb{P}$ denote the set of all odd prime numbers and $Bbb{N}$ be the set of all natural numbers. Let, $2a,2b$ be two even numbers both greater than $4$.
Define, $A={(p,q)inBbb{P}timesBbb{P}:p+q=2a}$ and $B={(r,s)inBbb{P}timesBbb{P}:r+s=2b}$. Let, $X'={A,B}$
Define, $n:X'to Bbb{N}$ by, $$n(Y)=max_{(x,y)in Y}{x^2+y^2},forall Yin X'={A,B}$$
My claim is, $n(A)=n(B)implies A=B$
I have no proof as well as no counterexample for my claim. If anyone has any idea about how to prove this or have seen any paper on this or it is very trivial to prove or has a counterexample for this claim please give.
Thanks in advance!
P.S.: I am "in danger of being blocked from asking any more".
number-theory elementary-number-theory prime-numbers diophantine-equations
edited Dec 7 at 15:44
Asaf Karagila♦
301k32423755
asked Dec 7 at 15:32
Sujit Bhattacharyya
945317
|
show 4 more comments
Let, $Bbb{P}$ denote the set of all odd prime numbers and $Bbb{N}$ be the set of all natural numbers. Let, $2a,2b$ be two even numbers both greater than $4$.
Define, $A={(p,q)inBbb{P}timesBbb{P}:p+q=2a}$ and $B={(r,s)inBbb{P}timesBbb{P}:r+s=2b}$. Let, $X'={A,B}$
Define, $n:X'to Bbb{N}$ by, $$n(Y)=max_{(x,y)in Y}{x^2+y^2},forall Yin X'={A,B}$$
My claim is, $n(A)=n(B)implies A=B$
I have no proof as well as no counterexample for my claim. If anyone has any idea about how to prove this or have seen any paper on this or it is very trivial to prove or has a counterexample for this claim please give.
Thanks in advance!
P.S.: I am "in danger of being blocked from asking any more".
number-theory elementary-number-theory prime-numbers diophantine-equations
edited Dec 7 at 15:44
Asaf Karagila♦
301k32423755
asked Dec 7 at 15:32
Sujit Bhattacharyya
945317
the notation is a bit unnecessarily complicated and obtuse. The entire three paragraph description can be simplified to, "fix even $a,binmathbb Z^+$ each greater than 4. If the sum of the squares of two primes whose sum is each of $a,b$ are equal, are $a$ and $b$ equal?"
– YiFan
Dec 7 at 16:37
If you have such kinds of conjectures you should write a program that tests this for a lot of numbers. Testing for all n<1000 should not be a problem.
– miracle173
Dec 7 at 17:38
1
can you give an example of your claim?
– miracle173
Dec 7 at 18:05
@YiFan Thank you for your kind response. But the statement you made in the comment is a bit wrong. Take $a=31b=7$ and $p=29,q=13$ then $p^2+q^2=a^2+b^2=1010$ but $pne a,qne b$. That's why I take the maximum of them.
– Sujit Bhattacharyya
Dec 8 at 2:31
@miracle173 I am having same trouble with the examples. Assuming $n(A)=n(B)implies max{a^2+b^2}=max{p^2+q^2}implies a^2+b^2=p^2+q^2$ for some $(a,b)in A,(p,q)in B$ where the maximum attained. From here if I am able to show that $a=p,b=q$ then we are done. But still no proof is found.
– Sujit Bhattacharyya
Dec 8 at 2:41
|
show 4 more comments
Let, $Bbb{P}$ denote the set of all odd prime numbers and $Bbb{N}$ be the set of all natural numbers. Let, $2a,2b$ be two even numbers both greater than $4$.
Define, $A={(p,q)inBbb{P}timesBbb{P}:p+q=2a}$ and $B={(r,s)inBbb{P}timesBbb{P}:r+s=2b}$. Let, $X'={A,B}$
Define, $n:X'to Bbb{N}$ by, $$n(Y)=max_{(x,y)in Y}{x^2+y^2},forall Yin X'={A,B}$$
My claim is, $n(A)=n(B)implies A=B$
I have no proof as well as no counterexample for my claim. If anyone has any idea about how to prove this or have seen any paper on this or it is very trivial to prove or has a counterexample for this claim please give.
Thanks in advance!
P.S.: I am "in danger of being blocked from asking any more".
number-theory elementary-number-theory prime-numbers diophantine-equations
edited Dec 7 at 15:44
Asaf Karagila♦
301k32423755
asked Dec 7 at 15:32
Sujit Bhattacharyya
945317
Let, $Bbb{P}$ denote the set of all odd prime numbers and $Bbb{N}$ be the set of all natural numbers. Let, $2a,2b$ be two even numbers both greater than $4$.
Define, $A={(p,q)inBbb{P}timesBbb{P}:p+q=2a}$ and $B={(r,s)inBbb{P}timesBbb{P}:r+s=2b}$. Let, $X'={A,B}$
Define, $n:X'to Bbb{N}$ by, $$n(Y)=max_{(x,y)in Y}{x^2+y^2},forall Yin X'={A,B}$$
My claim is, $n(A)=n(B)implies A=B$
I have no proof as well as no counterexample for my claim. If anyone has any idea about how to prove this or have seen any paper on this or it is very trivial to prove or has a counterexample for this claim please give.
Thanks in advance!
P.S.: I am "in danger of being blocked from asking any more".
number-theory elementary-number-theory prime-numbers diophantine-equations
number-theory elementary-number-theory prime-numbers diophantine-equations
edited Dec 7 at 15:44
Asaf Karagila♦
301k32423755
asked Dec 7 at 15:32
Sujit Bhattacharyya
945317
edited Dec 7 at 15:44
Asaf Karagila♦
301k32423755
asked Dec 7 at 15:32
Sujit Bhattacharyya
945317
edited Dec 7 at 15:44
Asaf Karagila♦
301k32423755
edited Dec 7 at 15:44
Asaf Karagila♦
301k32423755
edited Dec 7 at 15:44
Asaf Karagila♦
301k32423755
301k32423755
asked Dec 7 at 15:32
Sujit Bhattacharyya
945317
asked Dec 7 at 15:32
Sujit Bhattacharyya
945317
asked Dec 7 at 15:32
Sujit Bhattacharyya
945317
945317
the notation is a bit unnecessarily complicated and obtuse. The entire three paragraph description can be simplified to, "fix even $a,binmathbb Z^+$ each greater than 4. If the sum of the squares of two primes whose sum is each of $a,b$ are equal, are $a$ and $b$ equal?"
– YiFan
Dec 7 at 16:37
If you have such kinds of conjectures you should write a program that tests this for a lot of numbers. Testing for all n<1000 should not be a problem.
– miracle173
Dec 7 at 17:38
1
can you give an example of your claim?
– miracle173
Dec 7 at 18:05
@YiFan Thank you for your kind response. But the statement you made in the comment is a bit wrong. Take $a=31b=7$ and $p=29,q=13$ then $p^2+q^2=a^2+b^2=1010$ but $pne a,qne b$. That's why I take the maximum of them.
– Sujit Bhattacharyya
Dec 8 at 2:31
@miracle173 I am having same trouble with the examples. Assuming $n(A)=n(B)implies max{a^2+b^2}=max{p^2+q^2}implies a^2+b^2=p^2+q^2$ for some $(a,b)in A,(p,q)in B$ where the maximum attained. From here if I am able to show that $a=p,b=q$ then we are done. But still no proof is found.
– Sujit Bhattacharyya
Dec 8 at 2:41
|
show 4 more comments
the notation is a bit unnecessarily complicated and obtuse. The entire three paragraph description can be simplified to, "fix even $a,binmathbb Z^+$ each greater than 4. If the sum of the squares of two primes whose sum is each of $a,b$ are equal, are $a$ and $b$ equal?"
– YiFan
Dec 7 at 16:37
If you have such kinds of conjectures you should write a program that tests this for a lot of numbers. Testing for all n<1000 should not be a problem.
– miracle173
Dec 7 at 17:38
1
can you give an example of your claim?
– miracle173
Dec 7 at 18:05
@YiFan Thank you for your kind response. But the statement you made in the comment is a bit wrong. Take $a=31b=7$ and $p=29,q=13$ then $p^2+q^2=a^2+b^2=1010$ but $pne a,qne b$. That's why I take the maximum of them.
– Sujit Bhattacharyya
Dec 8 at 2:31
@miracle173 I am having same trouble with the examples. Assuming $n(A)=n(B)implies max{a^2+b^2}=max{p^2+q^2}implies a^2+b^2=p^2+q^2$ for some $(a,b)in A,(p,q)in B$ where the maximum attained. From here if I am able to show that $a=p,b=q$ then we are done. But still no proof is found.
– Sujit Bhattacharyya
Dec 8 at 2:41
the notation is a bit unnecessarily complicated and obtuse. The entire three paragraph description can be simplified to, "fix even $a,binmathbb Z^+$ each greater than 4. If the sum of the squares of two primes whose sum is each of $a,b$ are equal, are $a$ and $b$ equal?"
– YiFan
Dec 7 at 16:37
the notation is a bit unnecessarily complicated and obtuse. The entire three paragraph description can be simplified to, "fix even $a,binmathbb Z^+$ each greater than 4. If the sum of the squares of two primes whose sum is each of $a,b$ are equal, are $a$ and $b$ equal?"
– YiFan
Dec 7 at 16:37
If you have such kinds of conjectures you should write a program that tests this for a lot of numbers. Testing for all n<1000 should not be a problem.
– miracle173
Dec 7 at 17:38
If you have such kinds of conjectures you should write a program that tests this for a lot of numbers. Testing for all n<1000 should not be a problem.
– miracle173
Dec 7 at 17:38
1
1
can you give an example of your claim?
– miracle173
Dec 7 at 18:05
can you give an example of your claim?
– miracle173
Dec 7 at 18:05
@YiFan Thank you for your kind response. But the statement you made in the comment is a bit wrong. Take $a=31b=7$ and $p=29,q=13$ then $p^2+q^2=a^2+b^2=1010$ but $pne a,qne b$. That's why I take the maximum of them.
– Sujit Bhattacharyya
Dec 8 at 2:31
@YiFan Thank you for your kind response. But the statement you made in the comment is a bit wrong. Take $a=31b=7$ and $p=29,q=13$ then $p^2+q^2=a^2+b^2=1010$ but $pne a,qne b$. That's why I take the maximum of them.
– Sujit Bhattacharyya
Dec 8 at 2:31
@miracle173 I am having same trouble with the examples. Assuming $n(A)=n(B)implies max{a^2+b^2}=max{p^2+q^2}implies a^2+b^2=p^2+q^2$ for some $(a,b)in A,(p,q)in B$ where the maximum attained. From here if I am able to show that $a=p,b=q$ then we are done. But still no proof is found.
– Sujit Bhattacharyya
Dec 8 at 2:41
@miracle173 I am having same trouble with the examples. Assuming $n(A)=n(B)implies max{a^2+b^2}=max{p^2+q^2}implies a^2+b^2=p^2+q^2$ for some $(a,b)in A,(p,q)in B$ where the maximum attained. From here if I am able to show that $a=p,b=q$ then we are done. But still no proof is found.
– Sujit Bhattacharyya
Dec 8 at 2:41
|
show 4 more comments
1 Answer
1
active
oldest
votes
I'm afraid the question is not well defined; or at least, not proven to be well defined. In your definition for the sets $A$ and $B$, you essentially say they are the sets of all pairs of odd primes which sum to $2a$ and $2b$, which are even numbers. The Goldbach Conjecture asserts that every even number ($>4$) can be expressed as the sum of two odd primes, and this unfortunately hasn't been proven. Hence, if the Goldbach Conjecture happens to be false, your sets $A$, $B$ become null sets, and so the question becomes ill-defined and doesn't make sense; indeed, asserting the well-definition of $A$, $B$ would be asserting Goldbach! (This is of course unless you're willing to define $max(x^2+y^2)$ to mean something else in this case, in which case your conjecture comes down to asserting that there is exactly one even positive integer that can't be expressed as a sum of odd primes, an equally tough problem to prove.)
answered Dec 7 at 16:48
YiFan
2,3391421
If I redefine the functional $n$ as $n(emptyset)=0$ will it be then well defined?
– Sujit Bhattacharyya
Dec 8 at 2:38
@SujitBhattacharyya the question of whether it's well defined then comes down to asking whether there's only exactly one even number which cannot be expressed as the sum of two primes. The answer is likely "no", though of course it is probably as hard as Goldbach to prove this.
– YiFan
Dec 8 at 3:37
Thanks for the help. Can I assume that proving/assuming this problem is to be true directly proves Goldbach Conjecture?
– Sujit Bhattacharyya
Dec 8 at 5:55
@SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,Bneqemptyset$ in particular, then goldbach is true.
– YiFan
Dec 8 at 7:15
add a comment |
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I'm afraid the question is not well defined; or at least, not proven to be well defined. In your definition for the sets $A$ and $B$, you essentially say they are the sets of all pairs of odd primes which sum to $2a$ and $2b$, which are even numbers. The Goldbach Conjecture asserts that every even number ($>4$) can be expressed as the sum of two odd primes, and this unfortunately hasn't been proven. Hence, if the Goldbach Conjecture happens to be false, your sets $A$, $B$ become null sets, and so the question becomes ill-defined and doesn't make sense; indeed, asserting the well-definition of $A$, $B$ would be asserting Goldbach! (This is of course unless you're willing to define $max(x^2+y^2)$ to mean something else in this case, in which case your conjecture comes down to asserting that there is exactly one even positive integer that can't be expressed as a sum of odd primes, an equally tough problem to prove.)
answered Dec 7 at 16:48
YiFan
2,3391421
If I redefine the functional $n$ as $n(emptyset)=0$ will it be then well defined?
– Sujit Bhattacharyya
Dec 8 at 2:38
@SujitBhattacharyya the question of whether it's well defined then comes down to asking whether there's only exactly one even number which cannot be expressed as the sum of two primes. The answer is likely "no", though of course it is probably as hard as Goldbach to prove this.
– YiFan
Dec 8 at 3:37
Thanks for the help. Can I assume that proving/assuming this problem is to be true directly proves Goldbach Conjecture?
– Sujit Bhattacharyya
Dec 8 at 5:55
@SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,Bneqemptyset$ in particular, then goldbach is true.
– YiFan
Dec 8 at 7:15
add a comment |
I'm afraid the question is not well defined; or at least, not proven to be well defined. In your definition for the sets $A$ and $B$, you essentially say they are the sets of all pairs of odd primes which sum to $2a$ and $2b$, which are even numbers. The Goldbach Conjecture asserts that every even number ($>4$) can be expressed as the sum of two odd primes, and this unfortunately hasn't been proven. Hence, if the Goldbach Conjecture happens to be false, your sets $A$, $B$ become null sets, and so the question becomes ill-defined and doesn't make sense; indeed, asserting the well-definition of $A$, $B$ would be asserting Goldbach! (This is of course unless you're willing to define $max(x^2+y^2)$ to mean something else in this case, in which case your conjecture comes down to asserting that there is exactly one even positive integer that can't be expressed as a sum of odd primes, an equally tough problem to prove.)
answered Dec 7 at 16:48
YiFan
2,3391421
If I redefine the functional $n$ as $n(emptyset)=0$ will it be then well defined?
– Sujit Bhattacharyya
Dec 8 at 2:38
@SujitBhattacharyya the question of whether it's well defined then comes down to asking whether there's only exactly one even number which cannot be expressed as the sum of two primes. The answer is likely "no", though of course it is probably as hard as Goldbach to prove this.
– YiFan
Dec 8 at 3:37
Thanks for the help. Can I assume that proving/assuming this problem is to be true directly proves Goldbach Conjecture?
– Sujit Bhattacharyya
Dec 8 at 5:55
@SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,Bneqemptyset$ in particular, then goldbach is true.
– YiFan
Dec 8 at 7:15
add a comment |
I'm afraid the question is not well defined; or at least, not proven to be well defined. In your definition for the sets $A$ and $B$, you essentially say they are the sets of all pairs of odd primes which sum to $2a$ and $2b$, which are even numbers. The Goldbach Conjecture asserts that every even number ($>4$) can be expressed as the sum of two odd primes, and this unfortunately hasn't been proven. Hence, if the Goldbach Conjecture happens to be false, your sets $A$, $B$ become null sets, and so the question becomes ill-defined and doesn't make sense; indeed, asserting the well-definition of $A$, $B$ would be asserting Goldbach! (This is of course unless you're willing to define $max(x^2+y^2)$ to mean something else in this case, in which case your conjecture comes down to asserting that there is exactly one even positive integer that can't be expressed as a sum of odd primes, an equally tough problem to prove.)
answered Dec 7 at 16:48
YiFan
2,3391421
I'm afraid the question is not well defined; or at least, not proven to be well defined. In your definition for the sets $A$ and $B$, you essentially say they are the sets of all pairs of odd primes which sum to $2a$ and $2b$, which are even numbers. The Goldbach Conjecture asserts that every even number ($>4$) can be expressed as the sum of two odd primes, and this unfortunately hasn't been proven. Hence, if the Goldbach Conjecture happens to be false, your sets $A$, $B$ become null sets, and so the question becomes ill-defined and doesn't make sense; indeed, asserting the well-definition of $A$, $B$ would be asserting Goldbach! (This is of course unless you're willing to define $max(x^2+y^2)$ to mean something else in this case, in which case your conjecture comes down to asserting that there is exactly one even positive integer that can't be expressed as a sum of odd primes, an equally tough problem to prove.)
answered Dec 7 at 16:48
YiFan
2,3391421
answered Dec 7 at 16:48
YiFan
2,3391421
answered Dec 7 at 16:48
YiFan
2,3391421
answered Dec 7 at 16:48
YiFan
2,3391421
2,3391421
If I redefine the functional $n$ as $n(emptyset)=0$ will it be then well defined?
– Sujit Bhattacharyya
Dec 8 at 2:38
@SujitBhattacharyya the question of whether it's well defined then comes down to asking whether there's only exactly one even number which cannot be expressed as the sum of two primes. The answer is likely "no", though of course it is probably as hard as Goldbach to prove this.
– YiFan
Dec 8 at 3:37
Thanks for the help. Can I assume that proving/assuming this problem is to be true directly proves Goldbach Conjecture?
– Sujit Bhattacharyya
Dec 8 at 5:55
@SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,Bneqemptyset$ in particular, then goldbach is true.
– YiFan
Dec 8 at 7:15
add a comment |
If I redefine the functional $n$ as $n(emptyset)=0$ will it be then well defined?
– Sujit Bhattacharyya
Dec 8 at 2:38
@SujitBhattacharyya the question of whether it's well defined then comes down to asking whether there's only exactly one even number which cannot be expressed as the sum of two primes. The answer is likely "no", though of course it is probably as hard as Goldbach to prove this.
– YiFan
Dec 8 at 3:37
Thanks for the help. Can I assume that proving/assuming this problem is to be true directly proves Goldbach Conjecture?
– Sujit Bhattacharyya
Dec 8 at 5:55
@SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,Bneqemptyset$ in particular, then goldbach is true.
– YiFan
Dec 8 at 7:15
If I redefine the functional $n$ as $n(emptyset)=0$ will it be then well defined?
– Sujit Bhattacharyya
Dec 8 at 2:38
If I redefine the functional $n$ as $n(emptyset)=0$ will it be then well defined?
– Sujit Bhattacharyya
Dec 8 at 2:38
@SujitBhattacharyya the question of whether it's well defined then comes down to asking whether there's only exactly one even number which cannot be expressed as the sum of two primes. The answer is likely "no", though of course it is probably as hard as Goldbach to prove this.
– YiFan
Dec 8 at 3:37
@SujitBhattacharyya the question of whether it's well defined then comes down to asking whether there's only exactly one even number which cannot be expressed as the sum of two primes. The answer is likely "no", though of course it is probably as hard as Goldbach to prove this.
– YiFan
Dec 8 at 3:37
Thanks for the help. Can I assume that proving/assuming this problem is to be true directly proves Goldbach Conjecture?
– Sujit Bhattacharyya
Dec 8 at 5:55
Thanks for the help. Can I assume that proving/assuming this problem is to be true directly proves Goldbach Conjecture?
– Sujit Bhattacharyya
Dec 8 at 5:55
@SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,Bneqemptyset$ in particular, then goldbach is true.
– YiFan
Dec 8 at 7:15
@SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,Bneqemptyset$ in particular, then goldbach is true.
– YiFan
Dec 8 at 7:15
add a comment |
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the notation is a bit unnecessarily complicated and obtuse. The entire three paragraph description can be simplified to, "fix even $a,binmathbb Z^+$ each greater than 4. If the sum of the squares of two primes whose sum is each of $a,b$ are equal, are $a$ and $b$ equal?"
– YiFan
Dec 7 at 16:37
If you have such kinds of conjectures you should write a program that tests this for a lot of numbers. Testing for all n<1000 should not be a problem.
– miracle173
Dec 7 at 17:38
1
can you give an example of your claim?
– miracle173
Dec 7 at 18:05
@YiFan Thank you for your kind response. But the statement you made in the comment is a bit wrong. Take $a=31b=7$ and $p=29,q=13$ then $p^2+q^2=a^2+b^2=1010$ but $pne a,qne b$. That's why I take the maximum of them.
– Sujit Bhattacharyya
Dec 8 at 2:31
@miracle173 I am having same trouble with the examples. Assuming $n(A)=n(B)implies max{a^2+b^2}=max{p^2+q^2}implies a^2+b^2=p^2+q^2$ for some $(a,b)in A,(p,q)in B$ where the maximum attained. From here if I am able to show that $a=p,b=q$ then we are done. But still no proof is found.
– Sujit Bhattacharyya
Dec 8 at 2:41