Why product of primes $+1$ is not divisible by any of the product primes?

Question

Given $n$ prime numbers, $p_1, p_2, p_3,\ldots,p_n$, then $p_1p_2p_3\cdots p_n+1$ is not divisible by any of the primes $p_i, i=1,2,3,\ldots,n.$ I dont understand why. Can somebody give me a hint or an Explanation ? Thanks.

Answer 1

Towards a contradiction, if some prime $p_i$ ($1\leq i\leq n$) divides $p_1p_2p_3\cdots p_n+1$, then because $p_i$ also divides $p_1p_2p_3\cdots p_n$, it follows that $p_i\mid 1$ (the difference), a contradiction.

Answer 2

First approach:

Let $P= 2\times3\times5\times7\times11\times13.$ Then:

The next number after $P$ that is divisible by $2$ is $P+2.$

The next number after $P$ that is divisible by $3$ is $P+3.$

The next number after $P$ that is divisible by $5$ is $P+5.$

The next number after $P$ that is divisible by $7$ is $P+7.$

The next number after $P$ that is divisible by $11$ is $P+11.$

The next number after $P$ that is divisible by $13$ is $P+13.$

So $P+1$ is not divisible by any of those.

Second approach:

If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $2$ then the quotient is $\underbrace{3\times5\times7\times11\times 13}_{\large\text{excluding 2}}$ and the remainder is $1.$

If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $3$ then the quotient is $\underbrace{2\times5\times7\times11\times 13}_{\large\text{excluding 3}}$ and the remainder is $1.$

If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $5$ then the quotient is $\underbrace{2\times3\times7\times11\times 13}_{\large\text{excluding 5}}$ and the remainder is $1.$

If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $7$ then the quotient is $\underbrace{2\times3\times5\times11\times 13}_{\large\text{excluding 7}}$ and the remainder is $1.$

If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $11$ then the quotient is $\underbrace{2\times3\times5\times7\times 13}_{\large\text{excluding 11}}$ and the remainder is $1.$

If $(2\times3\times5\times7\times11\times13) + 1$ is divided by $13$ then the quotient is $2\underbrace{\times3\times5\times7\times11}_{\large\text{excluding 13}}$ and the remainder is $1.$

(Appendix: $(2\times3\times5\times7\times11\times13) + 1 = 59\times509.$)

Answer 3

Suppose it divide bt $p_{i}$. Then $n \equiv 0 \mod p_{i}$.

But $n = p_{1} \dots p_{n} + 1 \equiv 1 \mod p_{i}$.

Answer 4

Simple example. Suppose I consider $2 \cdot 3 \cdot 5 \cdot 7 = 210$

Now, $2$ divides $210$, and so do $3$, $5$, and $7$ ... of course!

But what happens if you divide $210+1=211$ by $2$? You get a remainder of $1$ ... exactly because you got a remainder of $0$ when dividing $210$. And the exact same thing happens for $3$, $5$, and $7$

Answer 5

If some integer $m$ is divisible by e.g. $13$ then $m+1$ is not.

Now note that $m=p_1p_2\cdots p_n$ is divisible by every $p_i$ with $i\in\{1,\dots,n\}$ and draw conclusions.

Answer 6

Because if $k>1$ divides $n$ then $k$ can not divide $n+1$.

If $n = m*k$ then then $n + 1 = k(m + \frac 1m)$ and $m + \frac 1m$ is not an integer.

Or to put it another way. If $n = m*k$ then the next multiple of $k$ is $n+k = m*k + k$ which is larger than $n+1 = mk + 1$.

So because $p_i$ divides $p_1p_2.....p_n$ and $p_i > 1$ then $p_i$ can not divide $p_1p_2.....p_n + 1$ because $p_1p_2.....p_n + 1 = p_i(p_1.....p_{i-1}p_{i+1}... p_n + \frac 1{p_i})$ and $p_1.....p_{i-1}p_{i+1}... p_n + \frac 1{p_i}$ is not an integer.

!!!!VITALLY!!!! IMOPORTANT POST-SCRIPT: This does NOT mean $p_1p_2.....p_n + 1$ is prime! It just means that none of the $p_i$ for $i \le n$ divide it. It is possible that there is a $p_m; m > n$ that $p_m$ does divide $p_1p_2.....p_n + 1$. This is not a contradiction because $p_m$ does not divide $p_1p_2.....p_n$.

Answer 7

Simpler argument is that dividing by any $p_k$ we get the remainder $1$.

Answer 8

Let $$P = p_1p_2...p_n+1$$ and let $p$ be a prime such that $p\mid P$.

Then $p$ can not be any of $p_1,p_2,p_3,\cdots ,p_n$ otherwise $p$ would divide the difference $P-p_1p_2...p_n=1$ which is not possible.