Originally Posted by

**Moo** Hm...

Let a be a non-coprime number with 1729. Therefore, there is a common factor d, different from 1.

Is there b such as :

$\displaystyle a^b \equiv 1 [1729]$ ?

If so, we know that $\displaystyle a^b-1$ is a multiple of 1729 : $\displaystyle a^b-1=1729k$, $\displaystyle k \in \mathbb{Z}$

$\displaystyle a^b=1729k+1$

Since d divides a, it divides $\displaystyle a^b$. And it divides 1729 too. So it has to divide 1 too, which is not possible unless d=1 -> contradiction.

Is it better this way ? :)