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Does $\prod_{i=1}^{k}p_i=\sum_{i=1}^{k}p_i^2$ with $p_i$ distincts primes and $k\geq2$ have a solution ?

Here is what I already know :

  • There is no solutions if $k\equiv0\bmod2$ or if $k\equiv0\bmod3$.

  • It's a special case of Hurwitz equation ($A\prod_{i=1}^{k}p_i=\sum_{i=1}^{k}p_i^2$) with $A=1$ and a fondamental solution can't contain coprimes.

  • There is no solution for $k\leq12$.

This question is related (sub case) to this one

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BenLaz
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  • Related – Elaqqad Jul 30 '15 at 11:34
  • Because the only fondamental solution for $k=5$ and $A=1$ is (1,1,3,3,4). As it contains coprimes, this solutions will never generate a solution without coprimes. (new solutions are obtained by subtiting elements $p_x$ by ${A*\prod_{i=1}^{k}p_i\over p_x}-p_x$) – BenLaz Jul 31 '15 at 07:28
  • If the fondamental solution contains coprimes then there is no solution (for that fondamental solution) because all the "child solutions" of this fondamental solution will contain coprimes (in the case of (1,1,3,3,4), substitying a $3$ leads to (1,1,3,9,4) etc.) and we want a solution with distincts primes. – BenLaz Jul 31 '15 at 08:15
  • I see now the reason: if we have a solution containing two element which are not coprime then the generated solutions will always contain two elements which are not coprime. for example if we have $(a_0,b_0,c_0,\cdots)$ a fundamental solution with $d=\gcd(a_0,b_0)>1$ then always in the next solutions we will have $\gcd(a_n,b_n)>d$ (yes in a certain order) and this implies that there is no solution for the case $k=5$. (until now, I don't understand your explication). It would be helpful for other people to see your reasoning so I think it's better to include that (why k<12) in the question – Elaqqad Jul 31 '15 at 08:33

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