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The motivation for this question is from one of well known Landau Problems which asks for proof of the statement:

Are there infinitely many primes $p$ such that $p − 1$ is a perfect square? In other words: Are there infinitely many primes of the form $n^2 + 1$.

But this is not what I am asking. Before I ask my question let me explain the scenario.

A prime $p$ is called $\color{blue}{\text{Nice prime}} \ $if sum of its digits is of the form $n^2+1$ For example $37\rightarrow 3+7=3^2+1$ . The primes sum of whose digits is of the form $n^2$ (Eg $31\rightarrow 4=2^2$) or $n^2-1$ (Eg $71\rightarrow 8=3^2-1$) will be called Almost Nice primes.

The question is are there infinitely many Nice primes?

Now, I tried to find Nice and Almost Nice primes by hand till $400$ and here is what I've got:

Nice primes are:

$5,11,19,23,37,41,73,89,101,109,113,127,131,163,179,181,197,271,307,311,359,373$.

While Almost Nice primes are:

$n^2\rightarrow 1,3,31,79,97,103,211,277,367,349$

$n^2-1\rightarrow 3,17,53,71,107,233,251$

There is a reason why I called primes whose sum of digits is of the form $n^2$ and $n^2-1$ Almost nice primes.

If you have an Almost Nice Prime of the form $n^2-1$ then adding $2\times10^k$ to it (here $k$ is the highest power of $10$ in decimal expansion of $n^2-1$) will give you a Nice prime if it is a prime (by it I mean $n^2-1+2\times10^k$). In a similar manner, if a prime $p$ is an Almost Nice prime of the form $n^2$ then if $n^2+10^k$ is a prime then it will be a Nice prime.

But introducing Almost Nice prime is not much helpful as we have to make sure that $\text{Almost Nice prime (of the form $n^2$)}+10^k$ $\text{or Almost Nice prime (of the form $n^2-1$)+$2\times10^k$}$ is a prime before concluding that it met our Nice prime criteria.

Since the post is long, I again remind you the question. Are there infinitely many Nice primes?

Thanks.

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    $\begingroup$ What do you obtain with the prime number theorem for the expected number of primes $p \le x$ whose sum of digits is $N$ ? $\endgroup$ – reuns Sep 28 '17 at 18:12
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    $\begingroup$ Since $n^2+1$ is never divisible by $3$, it is even likely that for every number of the form $n^2+1$ with $n\ge 1$, there exists a prime $p$ with digit-sum $n^2+1$ because the number of numbers with the given digit sum, already for $n=8$ (there is a prime for $n=1,2,\cdots,7$) is so large that we can expect to find several primes. But to prove that (or even the weaker) statement will probably be very difficult, if not impossible. $\endgroup$ – Peter Sep 28 '17 at 18:35
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    $\begingroup$ Your list of Nice primes to 400 is not correct. You missed 2, and 373 is not a Nice prime. I recommend writing code to generate such things. Here is a list to 1000: 2,5,11,19,23,37,41,73,89,101,109,113,127,131,163,179,181,197,269,271,307,311,359,401,433,449,467,523,541,557,593,613,631,647,683,719,773,809,811,827,863,881,953,971. This sequence is not in the OEIS. $\endgroup$ – Matthew Conroy Sep 28 '17 at 18:35
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    $\begingroup$ It there are infinite many primes of the form $2\cdot 10^n+3$ or $4\cdot 10^n+1$ , then the statement is already true. $\endgroup$ – Peter Sep 28 '17 at 18:43
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    $\begingroup$ Thanks @MatthewConroy $\endgroup$ – Vidyanshu Mishra Sep 28 '17 at 18:46
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Even the set of positive integers with digit sum $101$, only having the digits $0$ and $1$ and ending with a $1$, contains $$\binom{n-2}{99}$$ $n$-digit numbers.

This means, that we have , for example, about $\large \color\red {10^{438}}$ numbers with a million digits in the set. Plenty of them should be primes, since they share no common factor.

If $n$ increases, the binomial coefficient grows much faster than $n$ itself. So there is an overwhelming statistical evidence that infinite many nice primes exist.

Of course, such an evidence is no proof.

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  • $\begingroup$ you made any progress?? Although incomplete but still it seems that your reasoning is the way to reach the conclusion. If not, should I place a bounty on the question (I am personally baffled in this matter so asking you)? $\endgroup$ – Vidyanshu Mishra Sep 29 '17 at 14:30
  • $\begingroup$ @VidyanshuMishra Placing a bounty is not a bad idea. Perhaps, someone knows a rigorous proof. $\endgroup$ – Peter Sep 29 '17 at 16:39
  • $\begingroup$ Ha ha, perhaps someone knows a rigorous proof and waiting for bounty...... $\endgroup$ – Vidyanshu Mishra Sep 29 '17 at 16:55

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