True of false: $N = p_1p_2\cdots p_k+1$ is prime for every positive integer $k$, where $p_1,p_2,\ldots ,p_k$ are the $k$ smallest prime numbers. We know primes are either of the form 4k+1 or 4k+3.
The products of 2 numbers of the form (4k+1) or (4k+3) is of the form 4k+1
The product of 2 numbers of the form (4k+1) and (4k+3) is of the form 4k+3
So case 1:
N = (4k+1) +1
= 4k + 2
= 2(k+1)
This is obviously not prime
Case 2:
N= (4k+3) +1
= 4k+4
= 4(k+1)
This is also obviously not prime.
So the question is false.
Is this correct? Why did the question include the fact that p1, p2, . . . , pk are the k smallest primes?
 A: $$
(2\cdot3\cdot5\cdot7\cdot11\cdot13)+1 = 59\cdot509,
$$
so
$(p_1\cdots p_k)+ 1$ is composite when $k=6$.
The question seems rather confused: the $k$th prime number is not of the form $4k+1$ or $4k+3$, but rather is $4\ell+1$ or $4\ell+3$ for some number $\ell$ that is different from $k$.  Or if you like, the $n$th (rather than $k$th) prime is of the form $4k+1$ or $4k+3$ where $k$ differs from $n$.
Then you seem to be trying to consider $p_k+1$ instead of $p_1\cdots p_k+1$.
A: Take a look at this http://primes.utm.edu/notes/proofs/infinite/euclids.html  It tells you  that what you think is a common mistake.
A: As the question says that they are smallest prime numbers $p_1=2$ , the statement that primes are of form $4k+1$ or $4k+3$ works only for odd primes. I think the correction goes this way the first case is $2(4k+1)+1=8k+3$ and the second case is $2(4k+3)+1=8k+7$. These can't be told to be prime or composite directly. This was the mistake. To disprove it the product of primes till $13 $ can be taken.
