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).

• How can you conclude that $\frac{Q-1}{R}$ is an integer without knowing what $p_a,p_b,$ etc. actually are? – Cameron Buie Apr 21 '13 at 15:36
• I'm sorry, this is invalid. I made a mistake and for a second I got confused myself and thought that $p1*p2*..$ is the square of $pa*pb*...$ and concluded that this is an integer, I forgot that $pa*pb*...$ is actually the square root of &p1*p2*...+1$. – Ovi Apr 21 '13 at 15:45 • Hint$\rm\ \ 2(2k+1) = (n-1)(n+1)\:\Rightarrow\:4\,$divides the RHS but not the LHS, contradiction. – Math Gems Apr 21 '13 at 15:52 • Somebody has already asked a similar question less than an hour ago, and received a very neat answer. – barak manos Feb 6 '15 at 17:21 2 Answers Your conclusion is wrong, because you are careless with your notation. Let me spell out the proper notation, which would indicate where you made a mistake. (Note that this does not invalidate that you proof might eventually lead to an answer, merely that your proof is currently incorrect.) Suppose we have$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. • The thing to learn from this for the OP: The square of an even number is always a multiple of$4$, the square of an odd number is always one more than a multiple of$8$(!) – Hagen von Eitzen Apr 21 '13 at 15:50 • @Calvin I don't see why$(N-1)$and$(N+1)$are both necessarily prime – Ovi Apr 21 '13 at 15:57 • @Ovi$(N-1)$and$(N+1)$aren't both necessarily prime, but Calvin's proof doesn't anywhere imply that they need to be; whether they're prime or not, it's still the case that$N^2-1 = (N-1)(N+1)$and that if one of$(N-1)$,$(N+1)$is even then so is the other. – Steven Stadnicki Apr 21 '13 at 16:36 • Oh I see If they are both even then$N^2-1$has a factor of 4 and there is at most a factor of 2 in$N^2-1\$. I thought Calvin was trying to say that N plus or minus 1 are prime and there is only 1 even prime. Well thanks for clearing my confusion – Ovi Apr 21 '13 at 17:54

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.