Is my proof correct regarding the non primality of 2⋅17a+1$2\cdot {17}^a+1$?

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⋅17a+1$2\cdot {17}^a+1$ where a∈N$a\in \mathbb{N}$.

Now, first of all, I tried to get some initial information about a$a$ and used MAPLE 14. I came to know that if a=47$a=47$ then the above number is indeed prime but otherwise for any a∈N∖{47}$a\in \mathbb{N}\mathrm{\setminus }\{47\}$ with a≤3000$a\le 3000$ so far, computationally it has been shown that 2⋅17a+1$2\cdot {17}^a+1$ is not prime.

So we can take the advantage of it and conjecture: ** No number of the form 2⋅17a+1$2\cdot {17}^a+1$ is prime if a≥3000$a\ge 3000$**

Proof:

Note that for a≥3000$a\ge 3000$, we have

provided a≡0[2]$a\equiv 0[2]$.

Next we check the case when a$a$ is odd. Assume that a=2a1+1,a1∈N$a=2a_1+1,a_1\in \mathbb{N}$. (when a=1$a=1$ then 2⋅171+1=35=5⋅7$2\cdot {17}^1+1=35=5\cdot 7$ hence not prime. So we see for a=3,5,7$a=3,5,7$ etc i.e. a=2a1+1$a=2a_1+1$ form.)

This time we see that

So next we have to check what happens if we let a1=2a2−1$a_1=2a_2-1$ form. In this case the number will become as

And then taking modulo 5, we obtain as

Thus we conclude that 2⋅17a+1$2\cdot {17}^a+1$ when a≥3000$a\ge 3000$, is composite number.

Please tell if I have made any mistake on proving this.

Thanking to all of you in advance.