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I just found this kind of “conjecture” recently. Define the set of positive prime numbers $P$. Suppose $$ n \in P, n\geq 3. $$ Define $S := \{1,p_1,\dots,p_{x-1}\}$ that are primes and $1$ which are smaller than $n$. (in which $p_{x-1}$ is the prime directly before n)

  1. Can you find at most 2 distinct $a, b \in S$ (it's like $n$ can be $c$ + $a$ whereas $a$ = 2 and $b$ could be $0$) that $a + b + c = n$ and $c = p_{x-1}$? Does this apply for all primes?
  2. Based on my observations, I can generalize the problem into this smaller problem. Let $g$ be the difference between $n$ and $p_{x-1}$, that is $$g := n - p_{x-1}.$$ Note that $g$ is always divisible by two (except the special case $n = 3$). So a smaller question is the following. Define $S_g := \{1,p_1,\dots,p_{y-1}\}$. Find $m, q \in S_g$ such that $m + q = g$ (or whether the number $1$ and every primes smaller than any positive even number $g$ form that positive even number $g$).The special case $g = 2$ has only $m = 2$ and $q = 0$.
  3. What if $n \in N$? Does this kind of “conjecture” also holds?
  4. Is there any conjectures or problems or questions or proofs... that has already been asked before and that are similar to mine? (I remember the Goldbach-Euler problem is a little bit similar to this).

Note: Based on computations on Goldbach conjecture, there are several even numbers that cannot be added by primes. Does $g$ ever fall on those counterexamples?

To give you some examples:

3 = 2 + 1

5 = 3 + 2

7 = 5 + 2

11 = 7 + 3 + 1

13 = 11 + 2

17 = 13 + 3 + 1

19 = 17 + 2

23 = 19 + 3 + 1

29 = 23 + 5 + 1

31 = 29 + 2

and so on. I have tested this for all primes up to 2029 by hands and the conjecture still holds.

Please answer this kind of silly question in a carefully way because I don't have much background of math.

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    $\begingroup$ This is the Goldbach conjecture, more or less. $\endgroup$
    – lulu
    Feb 16 at 14:47
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    $\begingroup$ Note: I don't suppose you meant the two uses of $n$ to be the same, right? There aren't going to be $n-1$ primes less than $n$ generally. I suggest changing one of the $n's$. $\endgroup$
    – lulu
    Feb 16 at 14:48
  • $\begingroup$ @DietrichBurde the OP specified $n≥3$. $\endgroup$
    – lulu
    Feb 16 at 14:54
  • $\begingroup$ So for $n=3$ we have $2$ primes less than $3$. In other words, $S=\{1,2,2\}$ ? $\endgroup$ Feb 16 at 14:55
  • $\begingroup$ No, it's kind of n = 3 -> S = {1,2} n = 5 -> S = {1, 2, 3} n = 7 -> S = {1, 2, 3, 5} $\endgroup$ Feb 16 at 14:56

1 Answer 1

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To summarize the discussion in the comments:

For $n\in \mathbb N$ (not restricting to primes) there are $\pi(n)$ primes less than or equal to $n$ (this is standard notation). This means there are $\pi(n-1)$ primes less than $n$.

The Goldbach conjecture would tell us that each even number ($>2$) is the sum of two primes. Thus, if $n>3$ were odd, $n-p_{\pi(n-1)}$ is even so Goldbach tells you want you want, at least in the case where $n-p_{\pi(n-1)}>2$. If you allow $a=b$, then you are ok even in the case $n-p_{\pi(n-1)}=2$ (as you can take $a=b=1$).

If $n$ is even, however, then $n-p_{\pi(n-1)}$ is odd so you'd need one of $a,b$ to be $2$, which would force $n-p_{\pi(n-1)}-2$ to be prime, which of course it usually is not. Thus your conjecture is generally false for even numbers. Of course, if you drop the condition that one of the summands be $p_{\pi(n-1)}$ you can apply Goldbach to $n-3$.

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  • $\begingroup$ $a, b$ are distinct. However, in the case that $n = 2 + p_{n-1}$, $a = 2$ and $b$ = 0 $\endgroup$ Feb 16 at 15:11
  • $\begingroup$ Ok, but you still have a major problem with even $n$ so long as you require that the largest prime be included as a summand. And, again, please edit your original post to fix the index notation (don't use $n$ to denote two different things). $\endgroup$
    – lulu
    Feb 16 at 15:13

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