Note that \[2^{33}-2^{19}-2^{17}-1 = 3^{3}\cdot 13\cdot 661\cdot 37021\] So, $3\cdot 661=1983$ may serve as the required factor.

Eum... may I ask how do you "note that" without using software?

It was not said that using math software is forbidden.

Ah ok. Of course it is not forbidden, but I think there also is a nice solution to find this factorisation handwise.

let $ x = 2^{5*2 + 1}, y = - 2^{5*1 + 1}, z = - 1$, then $ 10^3 < x + y + z = 1983 < 5*10^3$ and $ x^3 + y^3 + z^3 - 3xyz = 2^{33} - 2^{19} - 2^{17} - 1$. it is well known that $ x + y + z | x^3 + y^3 + z^3 - 3xyz$ bye paolo

bboypa wrote: it is well known that $ x + y + z | x^3 + y^3 + z^3 - 3xyz$ Hi bboypa (aka Jordan), how do you demonstrate this? Bye

GioMott ...

GioMott wrote: bboypa wrote: it is well known that $ x + y + z | x^3 + y^3 + z^3 - 3xyz$ Hi bboypa (aka Jordan), how do you demonstrate this? $ x^3 + y^3 + z^3 - 3xyz=(y+x+z)(x^2+y^2+z^2-xy-yz-xz)$

Alternately, the polynomial $ x^3 - (3yz)x + (y^3 + z^3)$ has the root $ x = -y-z$.

$2^{33}-2^{19}-2^{17}-1=2^{33}-2^{18}-1-3\cdot2^{17}$ $x^3 + y^3 + z^3 - 3xyz =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$