Prove p_{1}p_{2}...p_{n}+1 is Not Divisible by Any of These Primes

[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!

 

[Hurkyl]

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}.

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?

 

[BustedBreaks]

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?

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.

 

[Hurkyl]

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.

That sounds perfectly reasonable -- you want to write a proof that works simultaneously for each of the p_x'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, ..., p_k-1, p_n, p_k+1, ..., p_n-1, p_k​

 

[icystrike]

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!

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}<br /> 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.