3
$\begingroup$

I've devised a formula, which I tried for many numbers got the correct result.

for every $x\in \mathbb N$, with $x ≠ 4 + 7w$ ($w\in \mathbb{N}$),

then $30x - 1= p$ is a prime number.


Example: $30 \times 3 - 1 = 90 - 1 = 89,$ and one can check that $89$ is a prime number.

Can someone prove or disprove my thesis?

$\endgroup$

closed as off-topic by GNUSupporter 8964民主女神 地下教會, Isaac Browne, steven gregory, B. Mehta, user223391 May 1 '18 at 20:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Isaac Browne, steven gregory, B. Mehta, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Welcome to the site. What is the question here? $\endgroup$ – Mankind Apr 22 '18 at 8:58
  • 1
    $\begingroup$ It will be easy corollary from Dirichlet's theorem on Arithmetic progression that your sequence will contain same ratio of prime with just $\mathbb N$ $\endgroup$ – didgogns Apr 22 '18 at 9:05
2
$\begingroup$

Take $x=30$, so $$30x-1=(30+1)(30-1)=31\times 29$$ Further, can you construct a counterexample for any linear ‘prime formula’ (i.e. $ax+b=p$)? How about polynomial?

$\endgroup$
3
$\begingroup$

No your claim does not work. Let's try this one: $$30 \times 7 - 1 = 11 \times 19,$$ in a more general context you can see that: $$30 \times (11t+7) - 1 = 11 \times (30t+19).$$

Also if we add more another condition $t \overset{7}{\not\equiv}1$, then one can check that $11t+7$ is not congruent to $4$ modulo $7$.


One can show that for every polynomial with integral coefficients like $P(x)$, always there is a value $x$, such that $P(x)$ is not prime.


Maybe you will find the Chinese remainder theorem helpful. Chinese remainder theorem says that there is a unique integer modulo $7$, such that: $$30x-1 \overset{7}{\equiv}0,$$ you can check that this unique integer modulo $7$, can be chosen to be $4$, and all that you have done is to remove $7w+4$.

Also, all that I have done is to repeat the same process for the prime number $11$.

$\endgroup$
  • $\begingroup$ @rtybase I think that he means that: for every $x\neq 4+7w$, $30x-1$ is prime. Did he mean this? or I am mistaken? $\endgroup$ – Davood Khajehpour Apr 22 '18 at 9:13
  • $\begingroup$ Yes, you are right, I deleted my comment. Although, maybe, it is worth showing $11t+7 \ne 4+7w$? $\endgroup$ – rtybase Apr 22 '18 at 9:15
2
$\begingroup$

I want to rewrite this $x \neq 4 + 7w$ as $x \not\equiv 4 \pmod 7$. So then, if, say $x \equiv 1 \pmod 7$, then $30x - 1$ should be a prime number. Then $30x - 1$ gives us the sequence 29, 239, 449, 659, $869 = 11 \times 79$... oops.

EDIT: I wanted to get something in before the question got closed. I knew it would take me a little bit of time to come up with the full set of six numbers that I wanted. Not a lot of time, but I didn't know how much time I had left to squeak something in.

Anyway, the numbers are:

  • $30 \times 29 - 1 = 869 = 11 \times 79$
  • $30 \times 23 - 1 = 689 = 13 \times 53$
  • $30 \times 10 - 1 = 299 = 13 \times 23$
  • $30 \times 26 - 1 = 779 = 19 \times 41$
  • $30 \times 55 - 1 = 1649 = 17 \times 97$
  • $30 \times 7 - 1 = 209 = 11 \times 19$

You have to remember that primes are closely spaced among the small numbers. So if you randomly (or pseudorandomly) pick small numbers, chances are good they're going to be prime. But if you pick larger numbers, you have less of a chance of hitting a prime.

$\endgroup$
1
$\begingroup$

Another counter-example is $x=7q \ne 7w+4, q\geq1$ and $30x-1=120q-1$. For $$q=120\cdot z^2 \Rightarrow 30x-1=(120z-1)\cdot(120z+1)$$ $\forall z\in \mathbb{N}\setminus\{0\}$ for example, thus a composite.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.