# Product of first $n$-th prime power integers $+ 1$

I was just playing with prime numbers and then I accidentally found this pattern.

Let $p_1\cdot p_2\cdot p_3\cdots p_n$ is the product of first $n$-th prime power integers. Prove that: $p_1\cdot p_2\cdot p_3\cdots p_n+1$ is also a prime.

## 3 Answers

It is unclear what you mean with prime powers.

If a prime power is written as $n = p^k$, where $p$ is prime, then it is not true as

$$2 \times 3 \times 2^2 + 1 = 25 = 5 \times 5.$$

If a prime power is written as $n = p^k$, where $p$ is prime and $k>1$, then it is not true as

$$2^2 \times 2^3 + 1 = 33 = 3 \times 11.$$

If a prime power is written as $n = p^k$, where $p$ is prime and $k$ is fixed, then it is not true as

$$2^2 \times 3^2 \times 5^2 + 1 = 901 = 17 \times 53.$$

• Thanks for the response. By the way, in mathematics, a prime power is a positive integer power of a single prime number, according to Wiki And you didn't read the question very well 'cause I clearly mentioned that $p^1⋅p^2⋅p^3⋯p^n$ is the product of first $n$-th prime power integers like your first example(If a prime power is written as $n=p^k$, where $p$ is prime). I forgot to mention the resulting integer could also be a prime power integer as well as a prime number @johannesvalks – user18724 Aug 25 '15 at 3:56

$1+2\times3\times5\times7\times11\times13=30031=59\times509$

• By prime power integers I meant numbers like $4,8,9$ and etc. Prime power integers are prime numbers and also numbers like $p^n$, where $p$ is a prime and $n$ is any integer. @Alessandro – user18724 Aug 24 '15 at 17:05
• @user18724 But $4 \times 8 + 1 = 33 = 3 \times 11$... Not prime... – johannesvalks Aug 24 '15 at 17:25
• @user18724 it doesn't work for first powers of primes, as I showed in my answer, so your general statement about prime powers is also false. If you want a counterexample with an higher exponent $1+2^2\times3^2\times5^2=901=17\times53$ – Alessandro Codenotti Aug 24 '15 at 18:16

I'm not sure what a "prime power integer" is, as no prime can be written in the form $m^n$ with $m,n$ positive integers and $n>1$. Assuming you just mean to say that $p_1,p_2,\ldots,p_n$ are primes, this conjecture is false. What is true is that $p_1p_2\cdots p_n+1$ is relatively prime to $p_1,p_2,\ldots, p_n,$ which is what is needed in Euclid's proof that there are infinitely many primes.

• Except that Euclid's proof doesn't specify that it's the $n$ smallest primes; just that it's any arbitrary finite set of primes. If $S$ is any such set (e.g. $\{2,11,47\}$ then the prime factors of $1+\prod S$ are not members of $S$. ${}\qquad{}$ – Michael Hardy Aug 24 '15 at 17:04
• By prime power integers I meant numbers like $4,8,9$ and etc. Prime power integers are prime numbers and also numbers like $p^n$, where $p$ is a prime and $n$ is any integer. @MichaelHardy – user18724 Aug 24 '15 at 17:09
• @user18724 : I'm not the one who raised that question. ${}\qquad{}$ – Michael Hardy Aug 24 '15 at 17:13
• By prime power integers I meant numbers like $4,8,9$ and etc. Prime power integers are prime numbers and also numbers like $p^n$, where $p$ is a prime and $n$ is any integer. @Showhat – user18724 Aug 24 '15 at 17:21
• In that case, $2\cdot 3 \cdot 4+1=5^2$. – Showhat Aug 24 '15 at 17:22