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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 $n$?

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    $\begingroup$ The sequence of determinants is OEIS sequence A067276. Not that this helps... $\endgroup$ – Robert Israel Jul 12 '17 at 7:11
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    $\begingroup$ 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. $\endgroup$ – Robert Israel Jul 12 '17 at 15:03
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    $\begingroup$ Also with determinant $0$: $$ \pmatrix{2 & 3 & 5 & 7\cr 11 & 13 & 17 & 19\cr 23 & 29 & 31 & 37\cr 41 & 47 & 67 & 73\cr }$$ $\endgroup$ – Robert Israel Jul 12 '17 at 15:12
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    $\begingroup$ @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. $\endgroup$ – Rofl Ukulus Jul 12 '17 at 17:01
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    $\begingroup$ 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... $\endgroup$ – Robert Israel Jul 13 '17 at 14:34

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