Guessing-wise, it seems easier to find such pairs for XOR-Goldbach because you can XOR two larger numbers to get a smaller one. Doesn't hint at a proof, but it seems that the probability of not finding one feels smaller?

That is to say, for a counter-example to exist for an even number $$$n$$$, there would have to be a "forbidden zone" in the bit-patterns of all primes that, miraculously, perfectly aligns with the bits of $$$n$$$, which feels very unlikely.

Well, note that if one chooses $$$k$$$ with $$$2^k \gt n$$$, two primes $$$p, q$$$ have XOR equal to $$$n$$$ if and only if they are expressible as $$$p = C \cdot 2^{k} + a$$$ and $$$q = C \cdot 2^{k} + b$$$ for odd integers $$$a, b$$$ satisfying $$$a \oplus b n$$$.

So then for a sufficiently large fixed $$$n$$$, the problem is equivalent to saying that given the linear polynomials

can we pick an $x$ so that one of the $$$2^{k-2}$$$ pairs or so formed by the XOR are both primes simultaneously. I think questions like this are still open, and the best we can do is guarantee a lower bound of $$$c(m)$$$ primes given $$$m$$$ linear polynomials like above, where $$$c(m)$$$ goes to $$$\infty$$$ with $$$m$$$, albeit very slowly. [to use this result directly without using further additive structure, one would require $$$c(m) \gt m/2$$$, which seems very hard and very open]

For some special $$$n$$$ maybe the problem would be more attackable if the pairs have better shape? I don't think it is easy in general though to settle this.

I think its more probable to find solution for higher n , considering the prime density.

This is most likely true because even if $$$\min(a, b) \gt n$$$, $$$a\oplus b = n$$$ can still happens. In fact, checking this for an arbitrary number $$$n$$$ seems hard enough as we have to pretty much try all $$$a$$$ or we have to have some form of proof specific to $$$n$$$.

Also you stated that $$$2$$$ cannot be the xor of 2 primes, but if we managed to have primes in the form of $$$4k + 1$$$ and $$$4k + 3$$$, they will xor to $$$2$$$, like $$$17 \oplus 19 = 2$$$, so it actually work. Maybe you actually meant $$$a, b \leq n$$$ or $$$\min(a, b) \leq n$$$ or some other constraint?

Actual proof is probably hard but we can have a probabilistic reasoning:

• Up to $$$n$$$, there are roughly $$$\frac{n}{\ln(n)}$$$ primes. So we can say that if we pick a random $$$1 \leq a \leq n$$$ then there the chance that $$$a$$$ will be a prime is $$$\frac{1}{\ln(n)}$$$.
• Similarly, $$$b = a \oplus n \leq 2n$$$, so the chance that $$$b$$$ will be a prime is roughly $$$\frac{1}{\ln(2n)}$$$.
• So assume the chance that $$$a$$$ and $$$b$$$ are both primes are independent (they are not but they also don't follow any particular pattern), then the chance that both $$$a$$$ and $$$b$$$ are primes are $$$\frac{1}{\ln(2n)\ln(n)}$$$.
• And because we can try all $$$n$$$ values of $$$a$$$, the chance that at least one of them work is very high.
• The expected amount of $$$a$$$ that works is $$$\frac{n}{\ln(2n)\ln(n)}$$$ which is strictly increasing from $$$n = 2$$$
• The probability that at least one $$$a$$$ works is $$$1 - (1 - \frac{1}{ln(2n)\ln(n)})^n$$$ which approach $$$1$$$ very quickly.

The probabilistic reasoning also offer an algorithm to find a pair of $$$a$$$ and $$$b$$$ that will work in $$$O(\ln(n)\ln(2n)P) = O(\log(n)^2P)$$$ where $$$P$$$ is the complexity of checking prime up to $$$2n$$$.

I also think there exists solution that satisfy $$$\max(a, b) \leq n$$$ as long as $$$n - 2^{\lfloor\log_2(n)\rfloor}$$$ are big enough by the above probabilistic reasoning.

Schinzel's hypothesis H implies this statement is true. However, this statement is likely open just like the regular Goldbach conjecture.

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is it new meta to get GM using llm and then farm contribution? you forgot to delete comments here. next time try harder dear friend from China.

Your third claim, which states that every odd integer n in an odd base k can be represented as the digit-wise sum of two primes without carry, can fail when k = 3 due to the restrictive nature of base 3 constraints.

Using a counterexample generator that targets the worst-case suffix of an odd integer n ending in long runs of digit 2, I found that with n = 7615689291988173557 (resulting in a base 3 suffix of ...2222222222222222222222222), there is no solution when checked against the first 2.1 million primes. (Possible for astronomical magnitudes, albeit density of solutions is nearly non-existent.) This is because in base 3, valid prime digits can only be {1, 2}. If your target digit t = 2, then the only carry-free decomposition is 1 + 1 = 2. 2 + 0 or 0 + 2 is invalid as it forces prime ≡ 0 (mod 3), and 1 + 2 = 0 is invalid since it is not 2.

Because of this, you have zero degrees of freedom. When t = 2, it forces the decomposition 1 + 1 at that position. A string of 25 2s forces both primes to match a specific digit pattern which is overwhelmingly likely to be composite due to fixed small divisors.