A number theory problem by Marcos Araujo

This is my soluction for the problem:

$A$ , $B$ and $C$ are the roots of $x^3 -(A + B + C)X^2 + (AB+AC+BC)x + ABC = 0$

So, when we make $x = A$ , $x = B$ and $x = C$ the sentence is true:

$A^3 -(A + B + C)A^2 + (AB+AC+BC)A - ABC = 0$ (I)

$B^3 -(A + B + C)B^2 + (AB+AC+BC)B - ABC = 0$ (II)

$C^3 -(A + B + C)C^2 + (AB+AC+BC)C - ABC = 0$ (III)

Adding (I) + (II) + (III) we get:

$(A^3+B^3+C^3) - 3ABC = (A + B + C)(A^2 + B^2 + C^2) + (A + B + C)(AB+AC+BC)$

$(A^3+B^3+C^3) - 3ABC = (A + B + C)(A^2+B^2+C^2 +AB + AC + BC)$

This is a well known factorization.

Now lets make:

$2^{33} = \left(2^{11}\right)^3$ ;

$-2^{19} =-2\times 2^{18}= -2^{18} -2^{18} = (-2^6)^3 - 2\times 2^{17}$

$-2^{17} = -2^{11}\times 2^6$

Now lets look the original expression:

$2^{33} - 2^{19} - 2^{17} - 1$

and lets re write it as:

$\left(2^{11}\right)^3 +(-2^6)^3 +(-1^3) - 3\times 2^{11}\times (-2^6) \times (-1)$

Making $A = 2^{11}; B = -2^6$ and $C = -1$ we get:

$\left(2^{11}\right)^3 +(-2^6)^3 +(-1^3) - 3\times2^{11}\times (-2^6) \times (-1)$ $= (2^{11} - 2^6 - 1)(A^2+B^2+C^2 -AB - AC - BC)$ $= 1983\times(A^2+B^2+C^2 -AB - AC - BC)$

So 1983 is the wanted divisor.

Thanks for all who shared and liked :-)

its factors are: 3^(3) * 13 * 661 * 37021

so the answer => 3*661=1983

1, 3, 9, 13, 27, 39, 117, 351, 661, 1983, 5949, 8593, 17847, 25779, 37021, 77337, 111063, 232011, 333189, 481273,