Number algebra calculus theory

The first equation can be re-written to be

$(1000A + 100B + 10C + A) + 10((10A + A) - (10B + C)) = 5555$

$1111A = 5555$

So $A = 5$

The second equation can also be re-written to become

$(1000A + 100B + 10C + A) - (1000A + 100C + 10B + A) = -90$

$90B - 90C = -90$

$B - C = -1$

$B = C - 1$

We already know that $C = A - B$ so we can substitute

$B = (A - B) - 1$

$A = 5$ so

$B = (5 - B) - 1$

$B = 4 - B$

$2B = 4$

$B = 2$

If we substitute the values into the equation for $C$ we get

$C = (5) - (2) = 3$

Now

$n = (5)^2 - (2)^2 + (3)^2 - (5)^2 = 5$

Next is the functions

$f_1(n) = \sqrt{(5)^3 + 3(5)^2 + 3(5) - 19} = 14$

For the second one we know that

$n = \sqrt[3]{x} - 10$

So

$5 + 10 = \sqrt[3]{x}$

$15^3 = x = 3375$

That means that

$f_2(n) = \frac{42(3375)}{54} = 2625$

Now the final part

$\overline{ABCA} - (f_1(n) + f_2(n)) = 5235 - 2639 = 2596$

The prime divisors of $2596$ are $2, 11 \text{ and } 59$

The largest of those is $59$ so the answer is $59$