Determinants of two matrices and their sum, entries are consecutive primes

Question

In general determinants are not additive (related question). Nevertheless, matrices $A$ and $B$ can be chosen so that $\det A+\det B=\det\left(A+B\right)$ (related answer).

For which positive integer $n$ can $2 n^{2}$ consecutive prime numbers form a solution of the equation $$ \det A + \det B = \det\left(A + B\right) $$ with two $n\times n$ matrices $A$, $B$?

$n=2$

A solution for $n=2$:

• $A=\begin{pmatrix}17 & 19\\23 & 29\end{pmatrix}$, $B=\begin{pmatrix}31 & 37\\41 & 43\end{pmatrix}$, $\det A=56$, $\det B=-184$, $\det\left(A+B\right)=-128$

There are many solutions for $n=2$ but I don't know if there are infinitely many. Even with the added restriction that the 8 primes must appear in increasing order, I found 110 solutions with entries below ${10}^7$, the first few are:

A = [[17, 19], [23, 29]], B = [[31, 37], [41, 43]], det A = 56, det B = -184, det (A+B) = -128
A = [[149, 151], [157, 163]], B = [[167, 173], [179, 181]], det A = 580, det B = -740, det (A+B) = -160
A = [[521, 523], [541, 547]], B = [[557, 563], [569, 571]], det A = 2044, det B = -2300, det (A+B) = -256
A = [[1277, 1279], [1283, 1289]], B = [[1291, 1297], [1301, 1303]], det A = 5096, det B = -5224, det (A+B) = -128
A = [[5437, 5441], [5443, 5449]], B = [[5471, 5477], [5479, 5483]], det A = 10850, det B = -10990, det (A+B) = -140
A = [[19853, 19861], [19867, 19889]], B = [[19891, 19913], [19919, 19927]], det A = 277830, det B = -279090, det (A+B) = -1260
A = [[103963, 103967], [103969, 103979]], B = [[103981, 103991], [103993, 103997]], det A = 623754, det B = -624006, det (A+B) = -252
A = [[113147, 113149], [113153, 113159]], B = [[113161, 113167], [113171, 113173]], det A = 452576, det B = -452704, det (A+B) = -128

Are there infinitely many solutions $A=\begin{pmatrix}a & b\\c & d\end{pmatrix}$, $B=\begin{pmatrix}e & f\\g & h\end{pmatrix}$ where $a<b<c<d<e<f<g<h$ are consecutive primes?

$n=3$

Solutions for $n=3$:

• $A=\begin{pmatrix}3 & 5 & 7\\29 & 37 & 41\\43 & 59 & 67\end{pmatrix}$, $B=\begin{pmatrix}11 & 13 & 17\\19 & 23 & 31\\47 & 53 & 61\end{pmatrix}$, $\det A=120$, $\det B=-24$, $\det\left(A+B\right)=96$
• $A=\begin{pmatrix}2 & 3 & 5\\19 & 23 & 31\\41 & 53 & 61\end{pmatrix}$, $B=\begin{pmatrix}7 & 11 & 13\\17 & 29 & 37\\43 & 47 & 59\end{pmatrix}$, $\det A=176$, $\det B=448$, $\det\left(A+B\right)=624$

There is at least one solution formed from 18 consecutive primes with entries in increasing order first in $A$ and then in $B$:

$$A=\begin{pmatrix}726371 & 726377 & 726379\\726391 & 726413 & 726419\\726431 & 726457 & 726463\end{pmatrix},\quad B=\begin{pmatrix}726469 & 726487 & 726497\\726521 & 726527 & 726533\\726559 & 726589 & 726599\end{pmatrix}$$

$n=4$

A solution for $n=4$:

• $A=\begin{pmatrix}2 & 3 & 5 & 7\\11 & 13 & 17 & 19\\23 & 29 & 31 & 71\\79 & 89 & 107 & 127\end{pmatrix}$, $B=\begin{pmatrix}37 & 41 & 43 & 47\\53 & 59 & 61 & 67\\73 & 83 & 97 & 101\\103 & 109 & 113 & 131\end{pmatrix}$, $\det A=744$, $\det B=936$, $\det\left(A+B\right)=1680$

$n>4$

Are there solutions for all $n$? Are there solutions for all $n$ such that entries of each matrix are increasing row-by-row (as in the above examples)? What can be known about solutions?

Answer 1

Assuming standard number theory conjectures, there are infinitely many solutions for the $n=2$ case. More specifically, the prime $k$-tuples conjecture implies that there are infinitely many integers $p$ such that $$ \{p,p+2,p+6,p+12,p+14,p+20,p+24,p+26\} $$ comprise eight consecutive primes; and one can check that $$ \det\begin{pmatrix} p & p+2 \\ p+6 & p+12 \end{pmatrix} + \det\begin{pmatrix} p+14 & p+20 \\ p+24 & p+26 \end{pmatrix} = \det\begin{pmatrix} p+p+14 &p+2+ p+20 \\ p+6+p+24 & p+12+p+26 \end{pmatrix} $$ is an identity (for any real number $p$).

Answer 2

Here is a heuristic argument for there being infinitely many solutions for every $n$. (There is no hope of turning this into a proof.)

Fix $n$. Now for large $k$, let's try to estimate how many solutions there are with $p$ from $2^k - 2^{k-1}$ to $2_k + 2^{k+1}$

By the prime number theorem, there are $O(\frac{2^k}{k})$ primes in that interval. With odds bounded away from zero, the distance from $p_i$ to $p_{i+2n-1}$ is within $2nk = O(k)$. (Remember, $n$ is fixed and so is a constant.) Now $\det(A) + \det(B) - \det(A+B)$ is a a multinomial in the entries of degree $n$, with a complicated set of terms that is still $O(1)$ for fixed $n$. The terms from the first two determinants are all of the form $p_i + O(k)$, with $p_i$ being $O(2^k)$ The terms from the second two determinants are of the form $2p_i + O(k)$.

When we write out this polynomial, all terms are of the form $q_m(p_{i+1}-p_i, p_{i+2}-p_i, \ldots, p_{i+2n-1} - p_i)p_i^{n-m}$ or $r_m(p_{i+1}-p_i, p_{i+2}-p_i, \ldots, p_{i+2n-1} - p_i)2^{n-m}p_i^{n-m}$where $q_m$ and $r_m$ are multinomials of degree $m$. Also since a determinant has equal numbers of positive and negative terms, we know that $q_0 = r_0 = 0$.

Now $q_1$ has $O(k)$ possible values. $q_2$ has $O(k^2)$ possible values. And so on up to $q_{n}$ having $O(k^n)$ possible values. The same is true for $r_1$, $r_2$, and so on. (The constants are affected by the fact that there are $n!$ terms in a determinant. But that's just a biggish constant here.) Assuming that all of those polynomials are random within their range, the odds of all being zero simultaneously is $O(\frac{1}{k^{n^2 + 1})$.

But we have $O(\frac{2^k}{k})$ values of $p_i$ to consider. And therefore the number of solutions grows like $O(\frac{2^k}{\text{poly}(k)})$, which grows exponentially with $k$.

And therefore the sum of the count of solutions over all large $k$ must be infinite.

Answer 3

For which positive integer $n$ can $2 n^{2}$ consecutive prime numbers form a solution of the equation $$ \det A + \det B = \det\left(A + B\right) $$ with two $n\times n$ matrices $A$, $B$?

There are solutions with $n=5$, examples below. I think for each prime $p=2,3,\ldots,41$ there is at least one solution using 50 consecutive primes beginning at $p$.

• $A=\begin{pmatrix}2 & 3 & 5 & 7 & 11\\ 13 & 17 & 19 & 23 & 29\\ 31 & 37 & 41 & 43 & 47\\ 53 & 59 & 61 & 67 & 73\\ 139 & 163 & 179 & 199 & 223 \end{pmatrix}$, $B=\begin{pmatrix}71 & 79 & 83 & 89 & 97\\ 101 & 103 & 107 & 109 & 113\\ 127 & 131 & 137 & 149 & 151\\ 157 & 167 & 173 & 181 & 191\\ 193 & 197 & 211 & 227 & 229 \end{pmatrix}$, $\det A=5824$, $\det B=46080$, $\det\left(A+B\right)=51904$
• $A=\begin{pmatrix}3 & 5 & 7 & 11 & 13\\ 17 & 19 & 23 & 29 & 31\\ 37 & 41 & 43 & 47 & 53\\ 59 & 61 & 67 & 71 & 73\\ 103 & 179 & 193 & 223 & 227 \end{pmatrix}$, $B=\begin{pmatrix}79 & 83 & 89 & 97 & 101\\ 107 & 109 & 113 & 127 & 131\\ 137 & 139 & 149 & 151 & 157\\ 163 & 167 & 173 & 181 & 191\\ 197 & 199 & 211 & 229 & 233 \end{pmatrix}$, $\det A=35808$, $\det B=-38112$, $\det\left(A+B\right)=-2304$
• $A = \begin{pmatrix}5 & 7 & 11 & 13 & 17\\19 & 23 & 29 & 31 & 37\\41 & 43 & 47 & 53 & 59\\61 & 67 & 71 & 73 & 109\\139 & 157 & 181 & 199 & 233\end{pmatrix}$, $B = \begin{pmatrix}79 & 83 & 89 & 97 & 101\\103 & 107 & 113 & 127 & 131\\137 & 149 & 151 & 163 & 167\\173 & 179 & 191 & 193 & 197\\211 & 223 & 227 & 229 & 239\end{pmatrix}$, $\det A = 30976$, $\det B = -307200$, $\det\left(A+B\right) = -276224$
• $A = \begin{pmatrix}7 & 11 & 13 & 17 & 19\\23 & 29 & 31 & 37 & 41\\43 & 47 & 53 & 59 & 61\\67 & 71 & 73 & 79 & 83\\109 & 151 & 181 & 211 & 229\end{pmatrix}$, $B = \begin{pmatrix}89 & 97 & 101 & 103 & 107\\113 & 127 & 131 & 137 & 139\\149 & 157 & 163 & 167 & 173\\179 & 191 & 193 & 197 & 199\\223 & 227 & 233 & 239 & 241\end{pmatrix}$, $\det A = 4384$, $\det B = -47200$, $\det\left(A+B\right) = -42816$

There is a solution with $6\times 6$ matrices:

$A=\begin{pmatrix}2 & 3 & 5 & 7 & 11 & 13\\ 17 & 19 & 23 & 29 & 31 & 37\\ 41 & 43 & 47 & 53 & 59 & 61\\ 67 & 71 & 73 & 79 & 83 & 89\\ 97 & 101 & 103 & 107 & 109 & 113\\ 127 & 131 & 191 & 193 & 269 & 277 \end{pmatrix}$, $B=\begin{pmatrix}137 & 139 & 149 & 151 & 157 & 163\\ 167 & 173 & 179 & 181 & 197 & 199\\ 211 & 223 & 227 & 229 & 233 & 239\\ 241 & 251 & 257 & 263 & 271 & 281\\ 283 & 293 & 307 & 311 & 313 & 317\\ 331 & 337 & 347 & 349 & 353 & 359 \end{pmatrix}$, $\det A=-91200$, $\det B=-6804480$, $\det\left(A+B\right)=-6895680$