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Without using a computer prove that this Proth number cannot be a prime number :

$$43373\cdot 2^{49822}+1$$

2 Answers 2

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HINT $\rm\ c = (a\!+\!1,b\!+\!1)\ |\ a\: b^{2\:k}\! + 1\ $ by $\rm\:mod\ c\!\!:\ a,\!b\equiv {-}1\ \Rightarrow\ a\: b^{2\:k}\!+1 \equiv\: -\:(-1)^{2\:k}\!+1\equiv 0\:.$

Above $\rm\ \ c = (43374,3) = 3\:.\ $ Hence $\rm\ a\: b^{2\:k}\! + 1\:$ prime $\rm\:\Rightarrow\ a\!+\!1,\:b\!+\!1\:$ coprime (except in some trivial degenerate cases where the gcd $\rm\:c\:$ is not a proper factor).

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Reduce it modulo $3$. Recall $2\equiv -1$ and note $43373\equiv2$ modulo $3$, plus $49822$ is even.

  • 2
    Are you kidding... it's divisible by 3..2011-11-22
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    Is there some similar known criterion for all Proth numbers ?2011-11-22
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    @pedja: [Proth's theorem](http://en.wikipedia.org/wiki/Proth%27s_theorem), apparently.2011-11-22
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    For large numbers Proth theorem is useless without computer's help...2011-11-22
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    @pedja but proth's theorem is a practical way to test primality. Do you know a better algoritm ? As for large numbers , well even factoring is hard usually.2013-11-01