(Maximized) Determinant

The determinant of such matrix $P = \begin{pmatrix} 1&3&5 \\ -1&x&2 \\ 2&4&y \end{pmatrix}$ is

$\mathrm{det} (P) = 1(x)(y) + 3(2)(2) + 5(-1)(4) - 2(x)(5) - 4(2)(1) - y(-1)(3)$ $\mathrm{det} (P) = xy - 10x + 3y - 16$

while its permanent is

$\mathrm{perm} (P) = 1(x)(y) + 3(2)(2) + 5(-1)(4) + 2(x)(5) + 4(2)(1) + y(-1)(3)$

$\mathrm{perm} (P) = xy + 10x -3y$

since we only know that $\mathrm{perm}(P) + \mathrm{det}(P) = 4018$

we can add the two equations and get

$2xy - 16 = 4018$

$xy = 2017$

Now, knowing that $x$ and $y$ are strictly integers, it narrows down our options in finding the maximum value of its determinant, since $2017$ is prime. Specifically, we have four options: $(x,y) = (\pm 1, \pm 2017), (\pm 2017, \pm 1)$ .

Now, substituting these to our determinant equation, we get

$\mathrm{det}(P)_{x=1, y=2017} = 8042$

$\mathrm{det}(P)_{x=-1, y=-2017} = -4040$

$\mathrm{det}(P)_{x=2017, y=1} = -18166$

$\mathrm{det}(P)_{x=-2017, y=-1} = 22168$

So there, our maximum possible determinant for $P$ with the given conditions is $22168$ .