Is there a $3\times3$ matrix of distinct positive primes whose determinant is $0$?

I came across this while attempting a partial answer to a question here (which I forgot to favorite). However, I've made no headway on it.

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    $\begingroup$ A duplicate question is here. $\endgroup$ Feb 1, 2018 at 13:15
  • $\begingroup$ @DietrichBurde -- Since it's a duplicate, should it be removed? $\endgroup$
    – coffeemath
    Feb 1, 2018 at 13:20
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    $\begingroup$ I think, it is not an exact duplicate. I only wanted to give you back what you have "forgotten"(if it was this question). $\endgroup$ Feb 1, 2018 at 13:22
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    $\begingroup$ It's not a duplicate because that one requires that the primes be consecutive; this has no such restriction. But that question does prove existence of at least one solution (p=2213) $\endgroup$
    – smci
    Feb 2, 2018 at 8:37

2 Answers 2


Note that $$2\cdot3+5=11$$ $$2\cdot7+17=31$$ $$2\cdot19+29=67$$ Therefore a possible matrix that satisfies your conditions is $$\begin{bmatrix}3&7&19\\5&17&29\\11&31&67\end{bmatrix}$$ The rows are linearly dependent, twice the first row plus the second equalling the third. The matrix thus has determinant 0.

As mentioned by Mark Dickinson on the other answer, the Green–Tao theorem proves infinitely many matrices of distinct primes with determinant 0 at any size, the smallest such 3×3 example being $$\begin{bmatrix}3&5&7\\13&17&19\\23&29&31\end{bmatrix}$$ with twice the second row being the sum of the other two rows.


The extended Goldbach's conjecture implies that there are infinitely many such matrices. This conjecture implies that not only is every even number greater than 2 the sum of two primes but that large even numbers can all be written as the sum of two primes in many different ways.

Take any 3 distinct sufficiently large primes $p_1, p_2, p_3$. Then by Goldbach's conjecture we can find primes $q_i,r_i$ with $2*p_i = q_i + r_i$ for $i = 1,2,3$. By the extended Goldbach's conjecture, we can choose these primes so that all of the primes are distinct. Clearly the determinant of


is zero since twice the first row is the sum of the other two.

I don't know if you can prove the existence of infinitely many examples without using any unproven assumption.

On Edit In the comments @MarkDickinson gave a nice argument which shows that the Green-Tao theorem implies that there are infinitely many solutions, hence no unproven assumptions are needed. Furthermore, Mark's argument implies that there are infinitely many $n \times n$ solutions for any $n \geq 3$ (it is easy to see that no $2 \times 2$ solutions are possible). The Green-Tao theorem states that there are arbitrarily long arithmetic progressions of primes, which immediately implies that for any fixed $k$ there are infinitely many arithmetic progressions of length $k$.

If $n \geq 3$, pick $n^2$ primes in arithmetic progression. Arrange these column by column into an $n \times n$ matrix $A$. Then the rows are in arithmetic progression. If $n = 3$ the rows satisfy the equation $R_1 - 2R_2 + R_3 = 0$. If $n \geq 4$, the rows satisfy $R_1 - R_2 - R_3 + R_4 = 0$. Thus in all cases the rows are linearly dependent hence the determinant is 0.

This argument can be used to explicitly construct examples for $n \leq 5$, but no explicit arithmetic progression of 36 primes is currently known (according to Wikipedia, the current record is 26). An interesting programming problem would be to find large singular matrices of distinct primes. Note that taking $n^2$ primes in arithemtic progression is overkill. It would suffice to find $n$ disjoint arithmetic progressions of length $n$.

On further edit The full power of Green-Tao is not needed. You get infinitely many such matrices of size $3 \times 3$ or larger as soon as you have infinitely many disjoint arithmetic sequences of length 3. Such things can be used to induce a linear dependence among the first 3 rows, after which the remaining rows can be filled in arbitrarily (subject to the constraint that the primes be distinct). Here is a 5x5 example:

$$ \begin{bmatrix}3&11&13&19&29\\5&17&37&31&41\\7&23&61&43&53\\2&47&59&67&71\\73&79&83&89&97 \end{bmatrix} $$

Note that this $5 \times 5$ matrix consists of the first 25 prime numbers. You can swap out 2 for 101 if you don't like even primes. With very little evidence I conjecture that given any $n \geq 4$, you can arrange the first $n^2$ prime numbers into a singular $n \times n$ matrix. I have verified that it works out to at least $n = 100$.

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    $\begingroup$ Thanks to Green and Tao, we know that there are arbitrarily long arithmetic progressions of primes, and hence infinitely many arithmetic progressions of primes of length 3; it seems to me that using those arithmetic progressions for the columns should give infinitely many examples satisfying $2q_1 = p_1 + r_1$, $2q_2 = p_2 + r_2$, $2q_3 = p_3 + r_3$. (Or I guess more simply any AP of length 9, rearranged into a 3-by-3 matrix, would work.) $\endgroup$ Feb 1, 2018 at 19:49
  • $\begingroup$ @MarkDickinson I thought about arithmetic sequences but didn't quite see how to pull it off (though I was mostly thinking in terms of Dirichlet's theorem where they need not be consecutive). Your parenthetical suggestion about rearranging an AP of length 9 is clever. Perhaps you could write that up as an answer. $\endgroup$ Feb 1, 2018 at 19:55
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    $\begingroup$ For the same reason as using APs of length 3 in each column would work: twice the middle row would be the sum of the first row and the last row. (Or if you write the values across first, then twice the middle column would be the sum of the first column and the last column.) $\endgroup$ Feb 1, 2018 at 20:03
  • $\begingroup$ For example, taking the first 9 terms of the progression here, GP/Pari gives me 0 from matdet([199, 409, 619; 829, 1039, 1249; 1459, 1669, 1879]). $\endgroup$ Feb 1, 2018 at 20:06
  • $\begingroup$ @MarkDickinson you are right. That is a good observation that you can get infinitely many solutions that way. If Green and Tao is really needed, then it is a surprisingly deep result that infinitely many solutions exist. Note that your argument doesn't just work for 3x3, but can be extended to larger matrices. $\endgroup$ Feb 1, 2018 at 20:07

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