# 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, $$A=\{(p,q)\in\Bbb{P}\times\Bbb{P}:p+q=2a\}$$ and $$B=\{(r,s)\in\Bbb{P}\times\Bbb{P}:r+s=2b\}$$. Let, $$X'=\{A,B\}$$

Define, $$n:X'\to \Bbb{N}$$ by, $$n(Y)=\max_{(x,y)\in Y}\{x^2+y^2\},\forall Y\in X'=\{A,B\}$$ 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".

• the notation is a bit unnecessarily complicated and obtuse. The entire three paragraph description can be simplified to, "fix even $a,b\in\mathbb Z^+$ 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 '18 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 '18 at 17:38
• can you give an example of your claim? – miracle173 Dec 7 '18 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 $p\ne a,q\ne b$. That's why I take the maximum of them. – Sujit Bhattacharyya Dec 8 '18 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 '18 at 2:41

## 1 Answer

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.)

• If I redefine the functional $n$ as $n(\emptyset)=0$ will it be then well defined? – Sujit Bhattacharyya Dec 8 '18 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 '18 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 '18 at 5:55
• @SujitBhattacharyya yes, you're right. If the stated conjecture is true and $A,B\neq\emptyset$ in particular, then goldbach is true. – YiFan Dec 8 '18 at 7:15