Is my proof correct? $p_1p_2p_3\cdots p_n+1)$ cannot be the square of an integer Prove that $p_1p_2p_3\cdots p_n+1$, where $p_n$ is the $n^{th}$ prime, cannot be the square of an integer.  
Let $p_1p_2p_3\cdots p_n+1=Q$ and assume it is the square of an integer, so $\sqrt{p_1p_2p_3\cdots p_n+1}$ is an integer and can therefore be written as $p_ap_bp_cp_d\cdots$ and let $p_ap_bp_cp_d\cdots=R$. Now $Q/R$ should be equal to $R$, so we have $(p_1p_2p_3\cdots p_n+1)/(p_ap_bp_cp_d\cdots)$, or $I+1/(p_ap_bp_cp_d\cdots)$, where $I$ is an integer. However, $I+$ a fraction is clearly not an integer, so this is a contradiction. (Note: I know this can be done with modular arithmetic, but I haven't learned it yet). 
 A: I'm the asker reviewing this question a few years later...
Now it's easy: squares are $\equiv 0, 1 \pmod 4$. Since $p_1 p_2 \cdots p_n$ has only one factor of $2$, it is $\equiv \pm2 \pmod 4$, so $p_1 p_2 \cdots p_n + 1 \equiv 3 \pmod 4$, thus it cannot be a square.
A: Your conclusion is wrong, because you are careless with your notation. Let me spell out the proper notation, which would indicate where you made a mistake. (Note that this does not invalidate that you proof might eventually lead to an answer, merely that your proof is currently incorrect.)
Suppose we have $Q = 1 + \prod_{i=1}^M p_i = R^2$ for some $M, R$ positive integers.
Let $R= p_a ^ {r_a} \times p_b ^{r_b} \times \ldots \times p_k ^{r_k}$, where $p_i$ are arranged in increasing order, $r_i$ are the exponents. Moreover, let $ p_l \leq p_M$. Then, your conclusion is that
$$R = p_a ^ {r_a} \times p_b ^{r_b} \times \ldots \times p_k ^{r_k} = \frac{Q}{R} = \frac { 1 + \prod_{i=1}^M p_i} {p_a ^ {r_a} \times p_b ^{r_b} \times \ldots \times p_k ^{r_k}}  $$
Now, we may not easily pull out your integer value of $I$, because
1. $r_i$ need not be 1.
2. There could be some primes which are larger than $p_M$.   

How about parity instead of modulo arithmetic? Very similar concepts though.
Proof by contradiction. Suppose that $1+ \prod_{i=1}^k p_i = N^2$ for some $k, N$ positive integers.
Then, $N^2 -1 = (N-1)(N+1) = \prod_{i=1}^k p_i$.
Since $p_1 = 2$, the RHS is even, which implies that either $N-1$ or $N+1$ is even. In either case, it implies that the other term is also even, hence the LHS is a multiple of $2 \times 2 = 4$. Thus, we get that the product of primes (LHS) is a multiple of 4, which contradicts the fact that there is only 1 even prime.
