Proving $\prod_{i=1}^np_i+1$ is not a perfect square

Question

Let $m=\displaystyle{\prod_{i=1}^np_n}$ be the product of the first $n$ primes $(n>1)$. prove that $m+1$ cannot be a perfect square.

I think that the opposite it correct: $m+1$ is not a complete square iff $(m+1)^{\frac {p-1}{2}}\equiv -1\pmod p$ but for each prime in the first $n$ primes $\displaystyle(m+1)^{\frac {p-1}{2}}=\sum_{k=0}^p\binom{n}{k}m^k$ and since $p\mid m$, $(m+1)^{\frac {p-1}{2}}\equiv 1\pmod p$ which means that allegedly it's a complete square.

Am I right?

Answer 1

Every prime $p$ greater than $2$ is congruent to $1$ or $3 \pmod{4}$. Next, note that $3^2 \equiv 1 \pmod{4}$. Hence, $\prod_{i=2}^n p_i \equiv 1$ or $3 \pmod{4}$. Once we include the first prime, $2$, we get $\prod_{i=i}^n p_i \equiv 2 \pmod{4}$. Thus:

$$\prod_{i=i}^n p_i + 1 \equiv 3 \pmod{4}$$

This cannot be a perfect square since all perfect squares are equivalent to $0$ or $1 \pmod{4}$.

Answer 2

What you have shown is that $m+1$ is a square mod each of the first $n$ primes, because it is $\equiv 1$ mod each of these primes and $1$ is a square. But you haven't checked that $m+1$ is a square mod each prime.

(Remark that it is true that an integer is a square if and only if it is a square mod $p$ for every $p$, so your strategy could have worked, had you checked for all the primes. But this fact is less trivial than it may seem at first glance! See if you can prove it.)

Now, as for your problem, here is a hint: the product of the first $n$ primes is $\equiv 2$ mod $4$.