Is my proof correct regarding the non primality of $2\cdot 17^a +1$? Today I need your help to know if the proof I have provided below is correct or not. I want to prove that there is no prime of the form $2\cdot 17^a+1$ where $a\in \mathbb N$.
Now, first of all, I tried to get some initial information about $a$ and used MAPLE 14. I came to know that if $a=47$ then the above number is indeed prime but otherwise for any $a\in \mathbb N\backslash \{47\}$ with $a\leq 3000$ so far, computationally it has been shown that $2\cdot 17^a+1$ is not prime.
So we can take the advantage of it and conjecture: ** No number of the form $2\cdot 17^a+1$ is prime if $a\geq 3000$**
Proof:
Note that for $a\geq 3000$, we have 
\begin{align*}
       &2\cdot 17^a+1\\
\equiv &(-1)(-1)^a+1[3]\\
\equiv &(-1)^{a+1}+1[3]\\
\equiv &0[3]
\end{align*}
provided $a\equiv 0[2]$. 
Next we check the case when $a$ is odd. Assume that $a=2a_1+1, a_1\in \mathbb N$. (when $a=1$ then $2\cdot 17^1+1=35=5\cdot 7$ hence not prime. So we see for $a=3, 5, 7 $ etc i.e. $a=2a_1+1$ form.)
This time we see that \begin{align*}
  &2\cdot 17^a+1\\
= &2\cdot 17^{2a_1+1}+1\\
\equiv &2(2)^{2a_1+1}+1[5]\\
\equiv &2(2^2)^{a_1}2+1[5]\\
\equiv &(-1)(-1)^{a_1}+1[5]\\
\equiv &(-1)^{a_1+1}+1[5]\\
\equiv &0[5]
\end{align*} 
provided $a_1+1$ is odd viz $a_1$ is even.
So next we have to check what happens if we let $a_1=2a_2-1$ form. In this case the number will become as $$2\cdot 17^{2a_1-1}+1=2\cdot 17^{4a_2-3}+1.$$
And then taking modulo 5, we obtain as \begin{align*}
&2\cdot 17^{4a_2-3}+1\\
\equiv & 2(2)^{4a_2-3}+1[5]\\
\equiv & 2(2)^{4a_2}2^{-3}+1[5]\\
\equiv & 2(1)(2^3)^{-1}+1[5]\\
\equiv & 2(3)^{-1}+1[5]\\
\equiv & 2\cdot 2+1[5]\\
\equiv & 0[5]
\end{align*}
Thus we conclude that $2\cdot 17^a+1$ when $a\geq 3000$, is composite number.
Please tell if I have made any mistake on proving this.
Thanking to all of you in advance.

No, your second computation and third computation rule out exactly the same candidates, all the $a$ congruent to $1$ mod $4$. You have never adressed the case where $a \equiv 3 \pmod 4$
There is absolutely no hope of doing a similar computation ruling out all the $a$ congruent to $3$ mod $4$, because the case $a = 47$ gives out a prime (notice how $47 \equiv 3 \pmod 4$).
This means that a proof only using congruences would have to work modulo some $M \ge 47$ and would have to show that $a \equiv 47 \pmod M \implies f(a) \equiv 0 \pmod {f(47)}$ (this is the only way to rule out the case $47$ mod $M$). Seeing how $f(47)$ is huge, $M$ also has to be huge (it has to be a multiple of the order of $17$ modulo $f(47)$ ; nothing like your $4$), and then you have to find primes to rule out the other $M-1$ cases (which would be nothing short of a miracle)
