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Since $2+3=5$ I know a simle example of a triple of coprime integers $(a,b,c)$, such that $$a+b=c,$$ being each summand a product of distinct primes $a=2$, $b=3$ and $c=5$, and their product is a primorial $$abc=2\cdot3\cdot 5.$$ It is the primorial or order $3$, that is the positive integer $30$.

I believe that the following problem should be in the literaute but I've curiosity about its solution

Question. What about the triples of (coprime) integers $(a,b,c)$ being solutions of the equation $$\underbrace{\text{product of distinct primes}}_{a}+\underbrace{\text{product of distinct primes}}_{b}=\underbrace{\text{product of distinct primes}}_{c}$$ that satisfy the condition $$abc=\text{a primorial}?$$ What I am asking is if it is known that there exists a $K$ such that $\forall k\geq K$ there are no solution of our previous problem, o well in other case there are infinitely many solutions. Many thanks.

My belief is that should be few solutions. If this problem was in the literature refer the literature as an answer, and I try find the article.

Edit: I didn't know such sequence ,and I didn't calculate terms of such. Many thanks @RobertIsrael for your reference to OEIS, I add here thus the name of the author of the sequence Naohiro Nomoto (Sep 10 2000), and more terms were added by Carlos Rivera.

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  • $\begingroup$ so then the question is if there is only finite solutions to a+b=c such that a,b and c are squarefree pairwise coprime integers such that abc is a primorial? $\endgroup$ – Jorge Fernández Hidalgo Aug 7 '17 at 7:51
  • $\begingroup$ Many thanks @TacNayn for your words, since my english is bad. One can read my question also as: What primorials $N$ can you represent as a sum $a+b=c$ with $abc=N$? I am interested only in how many of those can you represent, if these are a finite number, or well you can show that there are infinitely many solutions. $\endgroup$ – user243301 Aug 7 '17 at 7:56
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    $\begingroup$ Note that you would not require that $a$, $b$ and $c$ are coprime since that's implied by the fact that $abc$ is primorial. In fact they must be pairwise coprime - otherwise $abc$ would be divisible by a square which a primorial can't. $\endgroup$ – skyking Aug 7 '17 at 8:19
  • $\begingroup$ You are right @skyking . Many thanks. $\endgroup$ – user243301 Aug 7 '17 at 15:49
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For order $3$: $2+3=5$.

For order $4$: $3 + 7 = 10$.

For order $5$: $2 + 33 = 35$.

For order $6$: $13 + 42 = 55$.

For order $7$: $11 + 210 = 221$.

For order $8$: $57 + 385 = 442$.

For order $9$: $119 + 1311 = 1430$.

For order $10$: $1495 + 1463 = 2958$.

For order $11$: $374 + 22971 = 23345$.

For order $13$: $1235 + 495726 = 496961$.

EDIT: See OEIS sequence A057035.

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  • $\begingroup$ how did you find these? Brute force or some heuristics trying special combinations? $\endgroup$ – Jorge Fernández Hidalgo Aug 7 '17 at 8:10
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    $\begingroup$ A rather brutish search. $\endgroup$ – Robert Israel Aug 7 '17 at 8:13

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