Is the product of first $n$ prime numbers $+1$ another prime number? [closed]

Question

Hi,

I know that the answer is no, yet I dont know how to prove it wrong. Finding a counterexample is not a good solution because it is a past written exam question with no calculators allowed. The first counterexample is about 30000... Is there a, simple preferred, solution to this problem? (This is asked in a CS discrete math exam)

The original problem is from Rosen Discrete Math, and as follows:

Prove or disprove that $p_1p_2 ... p_n+1$ is prime for every integer n, where $p_i$ is the ith smallest prime number.

Thanks in advance.

Answer 1

Here's a possible intended solution to show that $30031 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ is composite without factoring it. Recall the Fermat primality test: if $a^{n-1} \not \equiv 1 \bmod n$, then $n$ cannot be prime. It turns out that $2^{30030} \equiv 21335 \bmod 30031$, so $30031$ must indeed be composite. There is a well-known algorithm called binary exponentiation that is reasonably fast to implement by hand and that could conceivably be done on an exam. (I am not totally convinced that this would be faster than trial division until $p = 59$, though. And if you followed Leonid's suggestion in the comments your life would be even easier.)

Edit: Here is the solution by trial division. It suffices to check the primes from $17$ up. Note that the fact that we know the prime factorization of $30030$ helps a lot. All of the arithmetic necessary beyond what I wrote down was mental.

For $17$ write $30031 = 30 \cdot 77 \cdot 13 + 1 \equiv -4 \cdot 9 \cdot -4 + 1 \equiv -8 \bmod 17$.

For $19$ write $30031 \equiv -8 \cdot 1 \cdot -6 + 1 \equiv 11 \bmod 19$.

For $23$ write $30031 \equiv 7 \cdot 8 \cdot -10 + 1 \equiv -99 \bmod 23$.

For $29$ write $30031 \equiv 1 \cdot -10 \cdot 13 + 1 \equiv -100 \bmod 29$.

For $31$ write $30031 \equiv 30000 \bmod 31$.

For $37$ write $30031 \equiv -7 \cdot 3 \cdot 13 + 1 \equiv -13 \bmod 37$.

For $41$ write $30031 \equiv -11 \cdot 5 \cdot 13 + 1 \equiv -14 \cdot 13 + 1 \equiv -140 \bmod 41$.

For $43$ write $30031 \equiv -13 \cdot -9 \cdot 13 + 1 \equiv -26 \bmod 43$.

For $47$ write $30031 = 210 \cdot 143 + 1 \equiv 25 \cdot 2 + 1 \equiv 4 \bmod 47$.

For $53$ write $30031 \equiv -2 \cdot -16 + 1 \equiv 33 \bmod 53$.

Finally, for $59$ write $30031 \equiv 33 \cdot 25 + 1 \equiv 11 \cdot 16 + 1 \equiv 0 \bmod 59$.

Answer 2

I don't know a non-computational solution to this question, or even a computational solution that wouldn't make one annoyed not to be allowed to use a calculator or computer. I don't know of any theoretical reason why the statement is false, and there are similar questions involving Euclid sequences that remain open: see e.g. Problem 6 of

http://alpha.math.uga.edu/~pete/NT2009HW1.pdf

A "U" in front of a part of the problem means that it is unsolved.

I am a professional mathematician specializing in number theory who has thought at least a little bit about similar problems in the context of teaching advanced undergraduate number theory. (I am not the world's greatest problem solver, but some of the people who have already commented on this question and not given a solution meeting the criteria above are about as quick and clever as they come.) So I submit that if I cannot solve the question in a nice way, then it is not reasonable to be able to expect an undergraduate to do so on an in-class exam.