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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.

KON3
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    I was confused until I realized that 2.17 is "2 times 17" and not a decimal number with fractional digits ... – Martin R May 22 '14 at 15:14
  • I think its right , but i guess taking 2 cases of 2a+1 and 2a-1 is not needed , as they both are same only – avz2611 May 22 '14 at 15:16
  • @MartinR I apologise for that. Edited now. Hope it is now perfect. – KON3 May 22 '14 at 15:18
  • @user142634 Ya, thats true. Actually while doing so even I also thought that but just for the continuation of typing, I didn't change 2a+1 and 2a-1. Thanks for response. – KON3 May 22 '14 at 15:19
  • Perhaps you might find this post interesting as well. – Lucian May 23 '14 at 00:31
  • @Lucian Thank you so much. That was really a helpful "linked" information for me. – KON3 May 23 '14 at 01:07

1 Answers1

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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)

mercio
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