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Let $n$ be an integer and $p_1,\ldots,p_{n^2}$ be the first prime numbers. Writing them down in a matrix $$ \left(\begin{matrix} p_1 & p_2 & \cdots & p_n \\ p_{n+1} & p_{n+2} & \cdots & p_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \cdots & \cdots & \cdots & p_{n^2} \end{matrix} \right) $$ we can take the determinant. How to prove that determinant is not zero for every $n$?

Davide Giraudo
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Rofl Ukulus
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    The sequence of determinants is OEIS sequence A067276. Not that this helps... – Robert Israel Jul 12 '17 at 07:11
  • Read comments from https://math.stackexchange.com/questions/2047879/matrix-generated-by-prime-numbers, especially Peter's comments are valuable. – Widawensen Jul 12 '17 at 10:21
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    @Widawensen Oh, you already asked the questions. Did you try the flintc library for checking n > 201. Not sure if it can handle those big matrices. I strongly believe that the proof will be difficult, too. Maybe it would be interesting to study the prime factorization of those determinants. – Rofl Ukulus Jul 12 '17 at 10:34
  • @RoflUkulus I think that Peter is the best expert on prime numbers ( much better than me) , ask him when he will be available. BTW he has asked many other interesting questions about primes.. – Widawensen Jul 12 '17 at 10:38
  • No zeros up to $n=460$. – Robert Israel Jul 12 '17 at 14:28
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    I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$\pmatrix{2 & 3 & 5\cr 7 & 11 & 13\cr 19 & 23 & 97\cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity. – Robert Israel Jul 12 '17 at 15:03
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    Also with determinant $0$: $$ \pmatrix{2 & 3 & 5 & 7\cr 11 & 13 & 17 & 19\cr 23 & 29 & 31 & 37\cr 41 & 47 & 67 & 73\cr }$$ – Robert Israel Jul 12 '17 at 15:12
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    @Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely. – Rofl Ukulus Jul 12 '17 at 17:01
  • @RobertIsrael so now 460 is the limit. I wonder why exactly 460 in this case? – Widawensen Jul 13 '17 at 09:57
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    At $n= 460$ the determinant has $1001$ digits. I was making a b-file for sequence A067276, and the OEIS doesn't like numbers with more than $999$ digits. I could go further, but computations start to slow down... – Robert Israel Jul 13 '17 at 14:34
  • @RobertIsrael Very inspiring, I hope some day a limit $n= 1000$ will be achieved. – Widawensen Jul 13 '17 at 17:35
  • @RobertIsrael The number of digits for determinant seems to increase (or at least not to decrease - almost linear function) . Could it be possible to compute with your software also series with this number of digits? – Widawensen Jul 13 '17 at 18:09
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    From my point of there are two things to study: 1.) Are there any permutations $\sigma \in S_{n^2}$ that permutate the entries such that the determinant goes to zero? If so, what do they look like? 2.) What are the least gaps (least is not really defined properly), that primes can have, to produce determinant zero, like the matrices Robert posted, for example maybe is there any $k$ such that $p_1,p_{1+k},p_{2+k},\ldots$ determinant gets zero. – Rofl Ukulus Jul 16 '17 at 10:54
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    @RoflUkulus We can also to check whether starting series from $1$ gives similar results. So far it was checked up to $n=200$. – Widawensen Jul 17 '17 at 12:27
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    No answer: You might look at "Integral Matrices" by Morris Newman. In particular the "Smith Normal Form". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere. – rrogers Jul 18 '17 at 18:37
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    But I want to tell those interested that the bound that arrives is $n=845$ with $714025$ primes in the maximum matrix. – Boris Valderrama Jan 02 '19 at 12:55
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    $n=951$ and there isn't counterexample – Boris Valderrama Jan 02 '19 at 18:13
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    The absolute value of the determinant of the matrix is strictly crescent. maybe by recursivity – Boris Valderrama Jan 08 '19 at 13:11
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    @Elborito This is because the Perron-Frobenius Eigenvalue grows faster than linearly and is much larger than the other eigenvalues (which are rarely smaller than one). Here is a list of the eigenvalues for $n=14$: $7843.09, -168.956, -11.4222 + 2.69341 i, -11.4222 - 2.69341 i, -5.68197 + 5.79415 i, -5.68197 - 5.79415 i, 0.121105 + 6.20539 i, 0.121105 - 6.20539 i, 4.28891 + 4.0145 i, 4.28891 - 4.0145 i, 11.132, 7.30824, -4.67599, -1.51002$. I've also made a plot of the lowest singular value for $n=10,\ldots,100$. But I can't see any pattern. E.g. , at $n=28$ the lowest sing. value is $0.006$. – amsmath Aug 29 '19 at 00:34
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    @Boris: Although the absolute value increases most of the time, there are exceptions. For example with $n=57,58$ the determinants are about $2.14 \times 10^{89}$ and $1.25 \times 10^{89}$, respectively. There are more exceptions later. – Jukka Kohonen Aug 24 '21 at 18:50
  • @Gerald Did you misread "not zero" as "nonnegative"? – rschwieb Jun 19 '23 at 19:06
  • Has anyone thought of posting this on math overflow? – bb_823 Jun 19 '23 at 19:37
  • Is it possible to prove this by showing that inverse of given matrix always exists as there always exists trivial solutions to Ax = 0 equation?. Trivial solutions are only possible as the rows of the matrix will always be independent, as each element of matix is a distinct prime number. – Aziz Jul 07 '23 at 10:35

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