the maximum value for $\det M + \det N$ Let $$ and $$ be any two $4 × 4$ matrices with integer entries satisfying
$$MN=2\begin{pmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 1 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 
\end{pmatrix}$$
Then the maximum value for $\det M + \det N$
My attempt:-
Let Eigen values of $M$ are $a,b,c,d$ and eigen values of $N$ are $x,y,z,w$. We need to maximize the function $f(a,b,c,d,x,y,z,w)=abcd+xyzw$. We also know that determinant $MN=16$. That is $g(a,b,c,d,x,y,z,w)=abcdxyzw=16$.
Using Lagrange Multiplier, $\nabla f=\lambda \nabla g$
$ bcd=\lambda bcdxyzw \tag{1}$
$ acd=\lambda acdxyzw \tag{2}$
$ abd=\lambda abdxyzw \tag{3}$
$ abc=\lambda abcxyzw \tag{4}$
$ yzw=\lambda abcdyzw \tag{5}$
$ xzw=\lambda abcdxzw \tag{6}$
$ xyw=\lambda abcdxyw \tag{7}$
$ xyz=\lambda abcdxyz \tag{8}$
From these equations $xyzw=\frac{1}{\lambda},$
$abcd=\frac{1}{\lambda}$
From the constraint $xyzwabcd=\frac{1}{\lambda^2}=16\implies \lambda=\pm\frac{1}{4}$
Applying $\lambda=\frac{1}{4}$ in the equations $a*(1)$,b*(2),c*(3),d*(4),x*(5),y*(6),z*(7),w*(8). We get $abcd=4$ and $xyzw=4$. So, $f(a,b,c,d,x,y,z,w)=8$. But answer is given as 17. How it is possible? Where is my mistake? Is there any shorter way?
 A: Let $A$ be given by
$$
A=
\pmatrix{
1&0&0&1\\
0&1&1&0\\
0&0&1&0\\
0&0&0&1
}
$$
From the equation $MN=2A$, it follows that $\det(M)\det(N)=16$.

It's easily verified that if the product of two integers is $16$, their sum is at most $16+1=17$, hence $17$ is an upper bound for $\det(M)+\det(N)$.

But the upper bound of $17$ can be realized using $M=2A$ and $N=I_4$, hence $17$ is the actual maximum. 

As regards your attempt . . .

What allows you to assume that the eigenvalues of $M,N$ are real?

More inportantly, you are not making use of the fact that the entries of $M$ and $N$ are integers.  
A: Since $MN=U$ and $U$ is an invertible matrix, both $M$ and $N$ must be invertible. Thus, we can write $\det(M)=\det(U)/\det(N)$. Together with the facts that the determinant of a matrix with integer entries is integer, and $U$ has a positive integer determinant, we can formulate this problem as follows:
$$\max_{x\in \mathbb Z} \ x+\frac{\det(U)}{x} \quad \text{ subject to } \quad 1\le x\le det(U).$$
The optimal value of this problem is simply equal to $\max\Big(\det(U)+1,2\sqrt{\det(U)}\textbf{1}_{\sqrt{U}\in \mathbb Z}\Big)=\det(U)+1$, which is obtained from $x=\det(U)$. 
