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I would like to accelerate the search of prime factors of

$$2\uparrow \uparrow 4+3\uparrow \uparrow 4$$

In a question, I asked for a prime factor and another user also asked, whether this number is prime.

To accelerate the search, I want to use the fact that $-65536$ is a cubic residue for every prime factor $p$ and even a $9$-th-power-residue. That means, the congruence $x^3\equiv -65536\ (\ mod\ p\ )$ and even the congruence $x^9\equiv -65536\ (\ mod\ p\ )$ is solveable, if $p$ is a prime factor of the given number.

This should restrict the possible factors much more than the condition $(\frac{-3}{p})=1$ ($-3$ is a quadratic residue mod $p$), but I do not have any idea, how the restrictions look like.

Peter
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    -3 being a quadratic residue implies that the prime should be in ${-1,1,7}$ mod 12. But the given number is 7 mod 12, which means that at least one prime factor should be 7 mod 12. You can search for that, though it may not be the smallest. – Aravind Nov 25 '15 at 15:28
  • Also $-(2^{16})$ being a cube is equivalent to 2 being a cube, since $-(2^{15})$ is a cube. – Aravind Nov 25 '15 at 15:31
  • https://github.com/gnufinder/prime-factor – Peter May 27 '16 at 06:10
  • http://programmers.stackexchange.com/questions/319713/can-i-find-the-remainder-more-efficiently/319715?noredirect=1#comment678647_319715 – Peter May 29 '16 at 12:53

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