Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Definition: Proth number is a number of the form :

$$k\cdot 2^n+1$$

where $k$ is an odd positive integer and $n$ is a positive integer such that : $2^n>k$

My question : If Proth number is prime number are there some other known relations in addition to $2^n>k$ , between exponent $n$ and coefficient $k$ ?

share|cite|improve this question
up vote 0 down vote accepted

There are many simple relationships that involve congruences. They have a flavour much like results you have mentioned in earlier posts.

For example, if $n>1$ is odd, then $k$ must be of the form $6a-1$ or $6a+3$. If $n$ is even, then $k$ must be of the form $6a+1$ or $6a+3$. The arguments are the familiar ones. For example, if $n$ is odd, then since $2\equiv -1\pmod 3$, it follows that $2^n\equiv (-1)^n =-1\pmod{3}$. But if $k$ is of the form $6a+1$, it follows that $k2^n \equiv -1 \pmod{3}$, and therefore $k2^n \equiv 0\pmod{3}$. If $n>1$, this means that $k2^n+1$ cannot be prime, since it is greater than $3$ and divisible by $3$.

We can obtain similar restrictions by working modulo primes greater than $3$. For example, suppose that $n \equiv 2\pmod {4}$, that is, $n$ is of the form $4a+2$. Then $2^n \equiv 4 \pmod {5}$, and therefore if $k \equiv 1\pmod {5}$, we have $k2^n +1\equiv 4+1=5 \pmod{5}$. This is impossible unless $n=2$ (and therefore $k=1$). So we conclude that if $n$ is of the form $4a+2$, and $k2^n+1$ is prime, then $k$ cannot be of the form $10b+1$ except in the case $n=2$, $k=1$. We can obtain similar restrictions on $k$ if $n$ is of the form $4a$, also $4a+1$, also $4a+3$.

Added: In a comment, the OP asks for a proof that if $k\equiv 1\pmod 3$ and $k\equiv 1\pmod{10}$ (and $k2^n+1$ is prime), then $n\equiv 0\pmod 4$. This is not absolutely true, take $n=2$, $k=1$, but let's not worry about isolated exceptions at the beginning. We need to rule out the other possibilities for $n$.

By the contents of the answer above, $n\equiv 2\pmod{4}$ is ruled out apart from a single exception. Now we need to rule out $n \equiv 1$ and $n\equiv 3$ (modulo $4$). That would make $n$ odd. In that case, again by the answer above, $2^n \equiv -1\pmod{3}$. So if $k \equiv 1\pmod{3}$, we find that $k2^n+1\equiv 0\pmod{3}$. With the single exception of $k=1$, $n=1$, this means that $k2^n+1$ cannot be prime.

share|cite|improve this answer
,I understand first part of your answer, for the second part : $11 \cdot 2^6 \not \equiv 5 \pmod 5 $,$704\equiv 4 \pmod 5$ – pedja Nov 18 '11 at 13:20
@pedja: Sorry, typo, it was supposed to be $k2^n+1 \equiv 4+1=5\pmod{5}$. And $k2^n+1$ is what we are worried about primality of. – André Nicolas Nov 18 '11 at 16:52
@pedja: I added some stuff at the end of the previous answer, since it was a bit long for a comment. The result follows fairly quickly from stuff that was already in the answer. – André Nicolas Nov 18 '11 at 18:36
,Yes I see,thanks for edit..I deleted comment because I have proved statement slightly different than you... – pedja Nov 18 '11 at 18:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.