Number Theory – Infinitely Many Primes of the Form p_3 = p_0 + p_1 + p_2

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Conjecture

There exists infinitely many primes of the form $p_3 = p_0 + p_1 + p_2$.

Proof

Assuming these conjectures are true:

Goldbach's conjecture Every even number greater than $2$ is the sum of two primes.

$$p_0 + p_1 = 2k$$

Maillet's conjecture Every even number is the difference of two primes.

$$p_3 – p_2 = 2k$$

For $k > 2$, equating the two expressions

$$p_0 + p_1 = p_3 – p_2$$

and rearranging

$$p_3 = p_0 + p_1 + p_2$$

Since there are infinitely many $k$, there are infinitely many primes of the form $p_3 = p_0 + p_1 + p_2$ $\blacksquare$

Examples

+-----+--------+----+-----+----+--------------------------+
|  k  |   2k   | p0 |  p1 | p2 |     p3 = p0 + p1 + p2    |
+-----+--------+----+-----+----+--------------------------+
|   5 |     10 |  3 |   7 |  3 |       13 = 3 + 7 + 3     |
|  50 |    100 |  3 |  97 |  7 |      107 = 3 + 97 + 7    |
| 250 |    500 | 13 | 487 | 23 |      523 = 13 + 487 + 23 |
+-----+--------+----+-----+----+--------------------------+

Best Answer

You can prove the same result on the strength of the Goldbach Conjecture alone. No need for Maillet's conjecture.

Let $p_0 \ge 7$ be any arbitrary prime number. Of course, any such number is odd, so $p_0 = 2k - 1$ for some positive integer $k \ge 4$. Now, minus the equation by $3$ to obtain

$$ p_0-3 = 2k-1-3 = 2k-4 = 2(k-2) $$

Hence, $p_0-3$ is twice some integer $k-2$, implying $p_0-3 \ge 4$ is even. Then, according to the Goldbach Conjecture, there exists two primes $p_1, p_2$ such that

$$ p_0-3 = p_1 + p_2 $$

and after adding $3$ to the equation, we have

$$ p_0 = p_1 + p_2 + 3 $$

Thus, if the Goldbach Conjecture is true, then every prime number $p_0 \ge 7$ is the sum of three primes: $3$ and the prime numbers $p_1, p_2$ such that $p_0-3 = p_1 + p_2$.

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