If $a_1,a_2,\ldots, a_n$ are distinct primes, and $a_1=2$, and $n>1$, then $a_1a_2\cdots a_n+1$ is of the form $4k+3$. If $a_1,a_2,\ldots, a_n$ are distinct primes, and $a_1=2$, and $n>1$, then $a_1a_2\cdots a_n+1$ is of the form $4k+3$.
I have no ideas how to approach this question. Any hints?
 A: If $a_1=2$ and $a_1,a_2,\ldots, a_n$ are distinct primes ($n> 1$), then $a_1a_2\cdots a_n+1$ is odd, so it's either of the form $4k+1$ or $4k+3$. For contradiction, assume it can be of the form $4k+1$. 
But then $a_1a_2\cdots a_n=4k$, so at least one of $a_2,\ldots, a_n$ must equal $2$, contradiction, because $a_1,a_2,\ldots, a_n$ are distinct (and $a_1=2$).
A: We have that $a_2,...,a_n$ are distinct from $a_1$ (which is $2$), so none of them is a multiple of $2$; hence $a_2\cdots a_n$ is odd, so $a_2 \cdots a_n -1$ is even (write it as $2k$). Then, look at:
$$a_1 \cdots a_n + 1= 2(2k+1) +1 = 4k +3$$
A: If $a_2, \dots,a_n$ are odd primes, they're congruent to $1$ or $3$ modulo $4$, and so is their product. Thus 
$$2a_2\dotsm a_n+1\equiv\begin{cases}2\cdot 1+1\equiv 3 \\2\cdot 3+1\equiv 3
\end{cases}\mod 4.$$
A: First forget that $3$ is a prime. Each odd prime apart from $3$ is of the form $6k\pm1$, and the product of such primes is of this form too. If we also multiply by $2$ and then add $1$ we get $12k\pm2+1$, that is either $12k+3$ or $12k-1$ clearly $\equiv 3\mod4$ (as $12\equiv0\mod4$ and $-1\equiv3\mod4$). 
Now if we take product of odd primes apart form $3$, it is of the form $6k\pm1$, and if we multiply by both $3$ and $2$, that is multiply by $6$ and then add $1$, the result is $36k\pm6+1\equiv((\pm6\mod4)+1)\equiv((2\mod4)+1)\equiv3\mod4$. 
