# On a conjecture and other numbers like $144$.

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?

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}

• I also wonder why the resilience to writing it in the form $100k^2+67$ rather than $k^2\Vert 67$. – user228113 Feb 13 '18 at 23:23
• 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. – Mark Bennet Feb 13 '18 at 23:24
• @MarkBennet yeah, but I introduced concatenation for users who wanted to learn something :) Also, thanks for that little tip – Mr Pie Feb 13 '18 at 23:26
• @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. – Mr Pie Feb 13 '18 at 23:28
• @RobertSoupe ok. I will look into that. Thank you :) – Mr Pie Feb 14 '18 at 2:27

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: $1,2916,5776,837225,3111696,10259209,18438436,24750625,63091249,$ $66879684, 67305616,67815225,71385601.$

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

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.