<s>[INST] Proving $\prod_{i=1}^np_i+1$ is not a perfect square 
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?
 [/INST] 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 $1 \pmod{4}$.
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