Let $A_{m \times n}$ is a matrix of real value. Denotes $a_{ij}$ is an element of $A$ at position $i,j$. The $f,g$ is a function for these elements. $$f(a_{ij})=(a_{ij}-m)^2-(a_{ij}-n)^2$$ and other equation such as $$g(a_{ij})=(a_{ij}-c)^2-(a_{ij}-d)^2$$
where $m,n,c,d$ are coefficients.
If given that $f(a_{ij})=g(a_{ij})$, could I conclude that $m=c$ and $n=d$? For the case $m=n$ and $c=d$ is so simple, hence, I did not mention here
Let $$ f(a) = (a - m)^2 - (a - n)^2 = a(2n - 2m) + n^2 - m^2$$ and $$ g(a) = (a - c)^2 - (a - d)^2 = a(2d - 2c) + d^2 - c^2.$$
If $f(a) = g(a)$ for all $a \in \mathbb{R}$ (or even if $f(a_1) = g(a_1)$ and $f(a_2) = g(a_2)$ for $a_1 \neq a_2$) then you can compare the coefficients and deduce that
$$ 2(n-m) = 2(d - c), \\ n^2 - m^2 = (n - m)(m + n) = (d - c)(d + c) = d^2 - c^2. $$
One solution is $n = m$ and $d = c$. If $m \neq n$ (and $d \neq c$) then you can divide by $m - n$ the second equation and obtain the equivalent system of equations
$$ n - m = d - c, \\ m + n = d + c $$
whose solution is $n = d$ and $m = c$.