Is there a number congruent to 1 modulo infinitely many primes?

Question

Let $A=\left\{ p_{r},p_{r+1},\dots\right\}$ a (infinte) set of consecutive prime numbers (if you prefer, if $\mathfrak{P}$ is the set of all prime numbers, $A=\mathfrak{P}-\left\{ 2,\dots,p_{r-1}\right\}$).

I want to show that doesn't exists a natural number $n$ such that $$n\equiv1\textrm{ mod }p_{i},\,\forall i=r,r+1,\dots$$

I think it's true but I'm note sure. Am I wrong?

Answer 1

If you allow $n=1$ your claim is false.

If $n>1$ is imposed, just note that there will be a prime $p_i$ in your list greater than $n$ and $n$ is certainly not $1$ modulo this prime.

Answer 2

Assuming that you want to exclude the trivial case $n=1$:

Hint: Since $A$ is an infinite subset of $\Bbb N$, $n$ is not an upper bound.

Another way: $n<p_n$ (if $p_n\notin A$ take $\min A$ instead of $p_n$).

Still another way: Take a prime divisor of $n!+1$. If it's not in $A$, take $\min A$.

Answer 3

Suppose $n>1$. Then $$p_r\cdot p_{r+1}\cdots \mid n-1$$ which means $$p_r\cdot p_{r+1}\cdots\leq n-1$$
The left hand side tends to infinity while the right hand side is a fixed number $n-1$. Bingo!