Take the 2-minute tour ×

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

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

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

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).

share|improve this answer
add comment

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

share|improve this answer
Are you kidding... it's divisible by 3.. –  Eric Naslund Nov 22 '11 at 19:02
Is there some similar known criterion for all Proth numbers ? –  pedja Nov 22 '11 at 19:09
@pedja: Proth's theorem, apparently. –  anon Nov 22 '11 at 19:19
For large numbers Proth theorem is useless without computer's help... –  pedja Nov 22 '11 at 19:36
@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. –  mick Nov 1 '13 at 13:46
add comment

Your Answer


