# Prime Number Proof Help.

1. Jan 30, 2010

### BustedBreaks

Let $$p_{1}, p_{2},...,p_{n}$$ be primes. Show that $$p_{1} p_{2}...p_{n}+1$$ is divisible by none of these primes.

Let $$p_{1}, p_{2},...,p_{n}$$ be primes
Let $$k \in N$$
Assume $$p_{1}p_{2}...p_{n}+1=kp_{n}$$

$$\frac{p_{1}p_{2}...p_{n}}{p_{n}}+\frac{1}{p_{n}}=k$$
$$p_{1}p_{2}...p_{n-1}+\frac{1}{p_{n}}=k$$

This is a contradiction because the left side will not be a natural number.

My issue is that this seems to only prove $$p_{1} p_{2}...p_{n}+1$$ is not divisible by $$p_{n}$$ and not all $$p_{1}, p_{2},...,p_{n}$$.

Thanks!

2. Jan 30, 2010

### Hurkyl

Staff Emeritus
That sounds right.

So, can you modify the proof to make it work for, say, p1?

Or, can you find a way to use what this argument proves to prove the general theorem?

3. Jan 30, 2010

### BustedBreaks

To make it work for $$p_{1}$$ I can just use $$p_{1}$$ instead of $$p_{n}$$, I guess. I don't know if that helps me see what to do...

I was thinking of writing something like what I had, but instead of $$p_{n}$$ on the right I would have $$p_{x}$$ where $$x\in{1,2,...,n}$$, but that seems weird.

4. Jan 30, 2010

### Hurkyl

Staff Emeritus
That sounds perfectly reasonable -- you want to write a proof that works simultaneously for each of the px's, so you have to use a variable.

Incidentally, the other method I was hinting at is substitution. Substitute into the original argument the list of primes
p1, ..., pk-1, pn, pk+1, ..., pn-1, pk

5. Jan 30, 2010

### icystrike

I guess you are doing the proof of infinite primes.(=

By using euclidean algorithm , you can conclude the gcd of $$p_{1}p_{2}...p_{n}+1$$ with $$p_{i}$$ for i=1,2,3...n will be 1 with a sound argument (hint: recall co-prime)

OR,

By using modular , $$p_{1} p_{2}...p_{n}+1\equiv1(modp_{i})$$ for i=1,2,3...n, recalling 1 is not a prime , you may draw a conclusion that suffices to prove your proposition.

Last edited: Jan 30, 2010