Doesn't this prove that $n^2+1$ contains infinitely many primes? The conjecture that $n^2+1$ contains infinitely many primes is equivalent to saying that $(n-1)^2+1$ contains infinitely many primes because the sequence $n-1$ contains all terms of the sequence $n$. 
$(n - 1)^2 + 1 = (n^2 - 2n + 1) + 1 = n^2 - 2n + 2 = n^2 + (-2n+2) = n * n + (-2n + 2)$.
Dirichlet tells us that any sequence $a + nd$ where $a$ and $d$ are coprime contains infinitely many primes. Now, consider the fact that $-2n + 2 = -2(n - 1)$ for any $n$ and that $n$ is always coprime with $n - 1$, which means that any odd $n$ is coprime with $-2(n - 1) = -2n + 2$. Since there are infinitely many odd numbers, $n * n + (-2n + 2)$ contains infinitely many numbers in the form $a + nd$ where $a$ and $d$ are coprime, which means that $n * n + (-2n + 2) = (n - 1)^2 + 1$ contains infinitely many primes, implying that $n^2+1$ also contains infinitely many primes.
 A: Let's have a look at the sequence $n\times n+(-2n+2)$
For example, if you put $n=9$, you will get $9 \times 9 +(-18+2)$
Now, we know that the sequence $9\times n+(-16)$ takes infinitely many prime values. However, $9 \times 9 + (-16)$ may or may not be one of those prime numbers. (in this case, it is not a prime)
Similarly, if you put $n=11$, you will get $11 \times 11 + (-22+2)$. Although the sequence $11 \times n + (-20)$ takes infinitely many prime values, $11 \times 11 +(-20)$ may or may not be one of those prime numbers. In this case, it is a prime.
Continuing this way, it is possible that you end up with only finitely many prime numbers.
A: You know, for example, that the sequence includes an infinite subset from at least one of the residue classes modulo $d$ for any $d$, including $d$ being a prime. That is simply a function of the sequence being infinite and there being only a finite number of residue classes.
But to use Dirichlet, you don't just need the numbers to be congruent modulo $d$, you need them to contain the whole arithmetic progression starting at some $a$.
