2
$\begingroup$

Yesterday, I noticed that the following numbers were prime: $$67, 167, 467, 967, 1667.$$ I realised that the digits of these primes, apart from $67$, were square numbers: $$0^2, 1^2, 2^2, 3^2, 4^2.$$

I then wanted to find values $a_n$ such that when we concatenate $a_n^{\ \ 2}$ with $67$, it is a prime number. This means, we join $a_n^{\ \ 2}$ with $67$ to form a new number, and we will denote this as $a_n^{\ \ 2}\mid\mid67$. For example, $1$ concatenated with $2$ is denoted as $1\mid\mid 2$ which equals $12$.

Below is a sequence of all the values $a_n$ that range from $0$ to $160$ inclusive:

$0,1,2,3,4,7,9,10,15,16,18,23,24,28,30,31,33,42,43,45,48,53,55,61,65,68,69,72,76,77,87,88,93,94,96,98,105,106,109,110,112,113,116,117,119,122,125,130,132,133,138,140,144,145,147,151,155,159.$

Here, $a_1 = 0$, $a_2 = 1$, $a_3 = 2$, $\ldots$

Conjecture 1. I made a conjecture that for every number $b_n^{\ \ 2}\mid\mid 67$ that is not prime, it will either have $2$ or $3$ divisors such that $b_n$ does not appear in this sequence. I did find a counter-example, namely $158$. This was the first counter-example thus far such that, $$158^2\mid\mid 67 = 24964\mid\mid 67 = 2496467 = 17\times 19\times 59\times 131.\tag{4 divisors}$$

Conjecture 2. I also made another conjecture that for $n\geqslant 3$, if $a_n^{\ \ 2}\mid\mid 67$ is prime, and $(a_n + 1)^2\mid\mid 67$ is also prime, then not only will $(a_n + 2)^2\mid\mid 67$ not be prime, but it will be semi-prime, which is a prime composed of strictly two prime numbers. However, I also found another counter-example, namely $144$. This was also the first counter-example to this conjecture, such that $144^2\mid\mid 67$ is prime, $(144^2 + 1)\mid\mid 67$ is also prime, but $$(144^2 + 2)\mid\mid 67 = 19\times 151\times 743.\tag{3 divisors}$$ This number is not semi-prime because it is composed of three prime numbers.

I extended my first conjecture for which every number $b_n^{\ \ 2}\mid\mid 67$ that is not prime will have at most $4$ prime divisors. But my question is, do there exist other counter-examples to the second conjecture? Particularly counter-examples that are square numbers, like $144$? And is there a counter-example to my extended first conjecture?

Thank you in advance.


Edit:

Conjecture 2 has been proven to be false, and lots of the counter-examples of conj$_2$ are square numbers. Here are the first few among the list: $$\begin{align} 1 = 1^2 &\quad 2916 = 54^2 &\quad 5776 = 76^2 &\quad \\ 837225 = 915^2 &\quad 3111696 = 1764^2 &\quad 10259209 = 3203^2\end{align}$$

$\endgroup$
8
  • 6
    $\begingroup$ I also wonder why the resilience to writing it in the form $100k^2+67$ rather than $k^2\Vert 67$. $\endgroup$ – user228113 Feb 13 '18 at 23:23
  • 3
    $\begingroup$ Is what you mean $100a_n^2+67$? That formulation is perhaps more amenable to analysis than your concatenation notation. It is easy to show that this is not divisible by any of $2, 3, 5, 7, 11, 13$. Avoiding low primes is a notorious way of creating conjectures about primes which are falsified for large values. $\endgroup$ – Mark Bennet Feb 13 '18 at 23:24
  • $\begingroup$ @MarkBennet yeah, but I introduced concatenation for users who wanted to learn something :) Also, thanks for that little tip $\endgroup$ – Mr Pie Feb 13 '18 at 23:26
  • 2
    $\begingroup$ @YuriyS I noticed that $67$ is prime, $167$ is prime, $467$ is prime, $967$ is prime, and $1667$ is prime. I found that the first few digits apart from $67$ were squared numbers. Then it came to this. $\endgroup$ – Mr Pie Feb 13 '18 at 23:28
  • 2
    $\begingroup$ @RobertSoupe ok. I will look into that. Thank you :) $\endgroup$ – Mr Pie Feb 14 '18 at 2:27
4
$\begingroup$

There are many $n$ such that $100n^2+67$, $100(n+1)^2+67$ and $100(n+2)^2+67$ are all prime. The sequence of such $n$ begins $$1,2,204,205,212,228,236,237,337,406,705,706,721,1016,1554,1765,1802,1879,1883,$$ $$1884,1945,1946,1954,1955,2088,2089,2184,2262,2294,2488,2744,2796,2858,2865,2914,$$ $$2915,2916,2940,2998,3076,3077,3408,3409,3434,3435,3664,3681,3707,3708,3810,3985,$$ $$4361,4616,4751,4752,5107,5549,5755,5776,5816,6146,6201,6223,6246,6247,6494,6809,$$ $$6858,7159,7476,7639,7697,7804,7912,8229,8230,8626,8908,8950,9137,9401,9628,\dots$$ Your conjecture 2 is false.

Added: Code to do this calculation in GP:

f(n)=100n^2+67

for(i=1,10^4,if(isprime(f(i))>0 && isprime(f(i+1))>0 && isprime(f(i+2))>0,print1(i,",")))

Added: Here are all the square terms less than 10^8: $1,2916,5776,837225,3111696,10259209,18438436,24750625,63091249,$ $66879684, 67305616,67815225,71385601.$

$\endgroup$
9
  • $\begingroup$ Wow!! How did you get all of those results so quickly?? $\endgroup$ – Mr Pie Feb 13 '18 at 23:39
  • 1
    $\begingroup$ Two lines of GP/PARI code. I'll add them to the answer. $\endgroup$ – Matthew Conroy Feb 13 '18 at 23:41
  • $\begingroup$ $2916=54^2$....... $\endgroup$ – Matthew Conroy Feb 13 '18 at 23:50
  • $\begingroup$ $837225 =915^2$ and $3111696=1764^2$ are in this list, too. I'm not sure this is relevant. $\endgroup$ – Matthew Conroy Feb 13 '18 at 23:51
  • $\begingroup$ Hahah yes you are right, but I was just curious. Thank you very much :) $\endgroup$ – Mr Pie Feb 13 '18 at 23:53
2
$\begingroup$

So you are looking at numbers of the form $a_k := 100k^2 + 67$. Let's see what are the conditions to have $p \ | \ a_k$.

$$ p \ | \ a_k \quad\Leftrightarrow\quad k^2 \equiv -100^{-1}\times 67 \pmod p. $$

Such $k$ exists if and only if $ \Big(\frac{-67}{p}\Big) = 1$. This happens at half of the primes. As there are infinitely many primes, you should be able to find $k$ using CRT such that $a_k$ has any number of divisors as you wish.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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