**A**lthough I am not receiving ** Le Monde** magazine by mail any longer (maybe a consequence of highly critical posts?!), I bought it at a newsagent last Saturday and found a fairly easy mathematical puzzle. Or is it that easy? The question is to decide about the unicity of the solution to the equation

when the matrix is *(p,p)*. When *p=9,10,11*, the matrix is invertible, so there is no problem. But, looking at *p=12*, it is no longer invertible! When running a quick check like

for (p in 3:30){ A=diag(p) A[(1:p)+p*(((1:p)+1)%%(p))]=1 print(p) b=try(solve(A))}

we actually spot that *p=4,8,12,16,20,24,28,…*, i.e. all multiples of 4, lead to a non-invertible matrix. It is easy to see why for the *(4,4)* matrix

since the first and second lines are repeated. The *(8,8)* matrix is equivalent to the bloc matrix

hence with the same determinant as … In the general case, the determinant of the matrix is a circular determinant. Because most entries (but two) are zero, the determinant can easily be written as

where *ι* is the basic complex number. Therefore, if, and only if, *p* is a multiple of 4, *j* in the above product can take the value *n/4*, leading to a zero in the product… I can see why this makes a good puzzle, because this case of the multiples of 4 is quite counterintuitive!