How would you show that not every number of the form $$N = (p_1 p_2 p_3 \cdots p_n) + 1$$ is prime, where $$p_1, p_2, p_3,...,p_n$$ is the list of all prime numbers?

I have tried several proof techniques including the proof of infinitely many primes but to no avail. Any suggestions would be appreciated.

• There is no exhaustive, yet finite list of prime numbers. Do you mean $p_1, \ldots, p_n$ are the first $n$ prime numbers? If so, I would recommend searching for a counterexample. – user754697 Mar 4 '20 at 10:29
• – Mauro ALLEGRANZA Mar 4 '20 at 10:33
• @user754697, $p_1$,...,$p_n$ is the list of all prime numbers. – DYBnor Mar 4 '20 at 10:33
• @DYBnor The point is that there are infinitely many primes. So either $p_1\times p_2\times \ldots \times p_n=2\times3\times\ldots$ is infinite or it contains a finite subset of the primes (e.g., the first ones). – Jam Mar 4 '20 at 10:35

You could just do it by counterexample. The Euclid numbers are the numbers of the form $$p_n\#+1$$, where $$p\#$$ denotes the "primorial" (the product of the primes less than or equal to $$p$$). These are exactly what you desire in form: $$p_n$$ denotes the $$n^{th}$$ prime in this context.
$$p_6\# +1 = 13\# + 1 = 30031 = 59 \times 509$$
• @ Evee Trainer, I am a bit confused. How can you use counterexample when the statement has got 'not' in it.? In am thinking in this case $p_6$# + 1 confirms the statement. – DYBnor Mar 4 '20 at 11:01
• @DYBnor It's a counterexample to the claim that all numbers of the form $p_n\#+1$ are prime. So it's an example that proves your claim. It just depends on which angle you look at the problem. Mostly, you'd phrase this type of conjecture as "proposition X is true for all $n$". – Jam Mar 4 '20 at 11:07