Verify my proof that $N$ in $N = (P_1 \cdot P_2…P_n)+1$ must be odd.

I'm a noob. Is the following valid?

$$N = (P_1 \cdot P_2...P_n)+1$$

where $Pi$ is the $i^{th}$ consecutive prime number and $n$ is any natural number. If $N$ is even, $\dfrac{N}{2} \in \Bbb{N}$, which can be put into the first equation like so:

$$\frac{N}{2} = \frac{(P_1 \cdot P_2...P_n)}{2}+\frac{1}{2}$$

where it is expected that $\dfrac{(P_1 \cdot P_2...P_n)}{2}+\dfrac{1}{2} \in \Bbb{N}$ also. $P_1 = 2$, being the first prime number and thus

$$\frac{(P_1 \cdot P_2...P_n)}{2} = \frac{2(P_2 \cdot P_3...P_n)}{2} = P_2 \cdot P_3...P_n$$

where $P_2 \cdot P_3...P_n \in \Bbb{N}$. Because $\dfrac{1}{2} \notin \Bbb{N}$, $\dfrac{(P_1 \cdot P_2...P_n)}{2}+\dfrac{1}{2} \notin \Bbb{N}$ because any natural number plus a non-natural number is not a natural number.

Because the RHS of the first equation does not belong to $\Bbb{N}$, $\dfrac{N}{2} \notin \mathbb N$ also. Thus $N$ is odd.

• Note that $N$ as you define it will always be odd since the product $(P_1 \cdot ... \cdot P_n)$ is necessarily even since it contains $P_1 = 2$. – Zubzub Apr 13 '17 at 9:35

That is really complicated, but yes, you are right. An easier way to see it would be the following:

As $P_1 = 2$, we have that $$P_1\cdot P_2 \cdot \ldots \cdot P_n = 2\cdot P_2 \cdot \ldots \cdot P_n$$ is even. Now $1$ is odd, and you might know that even plus odd always gives odd, case closed. :)

If $P_1 = 2$, obviously $N = 2k + 1$ is odd, so your argument after "If $N$ is even.." is invalid.

• No, it is not. He is assuming that $N$ is even and the proving by contradiction that this case can not be true. Thus the argument is valid, even though a proof by contradiction is maybe not the easiest choice here. – Dirk Apr 13 '17 at 9:50
• He edited the question to make it a proof by contradiction. The original post didn't conclude that $N$ was odd. – Aryaman Jal Apr 13 '17 at 9:51

This reminds me of Conway's PRIMEGAME, in which a list of fractions is used to obtain powers of 2 with prime indices. It makes the sieve of Eratosthenes look very sophisticated by comparison. But your question assumes we already know the prime numbers.

You state that $P_1 = 2$. That's obviously even. So is $P_1 P_2$. And $P_1 P_2 P_3$. And any $P_1 P_2 \ldots P_n$ for $n > 2$. These numbers are all divisible by 2 but not by 4: 2, 6, 30, 210, 2310, 30030, 510510, 9699690, 223092870, 6469693230, etc. See Sloane's A002110.

So halving $N$ is an entertaining but unnecessary sidetrack. Since $P_1 P_2 \ldots P_n$ is even, it immediately follows that $N = P_1 P_2 \ldots P_n + 1$ is odd.

This is not to say that there don't exist situations where halving a number is useful. There are such situations. It's just that this one is not such a situation.