Proving that $|d|$ is not a prime number

Assume $a,b,c$ are integers and $a+b+c=0$.

If $d=a^{1433}+b^{1433}+c^{1433},$ prove that $|d|$ is not a prime number.

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Note that $1433$ is prime. It follows by Fermat's theorem that $a^{1433}\equiv a\pmod{1433}$, with similar results for $b$ and $c$. Thus $$d=a^{1433}+b^{1433}+c^{1433}\equiv a+b+c\equiv 0\pmod{1433}.$$ We conclude that $|d|$ is divisible by $1433$. Finally, we need to show that $|d|$ cannot be equal to $1433$. This is essentially obvious, since $1433$-th powers of anything other than $0$, $1$, and $-1$ are very far apart.
Added: Using Fermat's Theorem was ludicrous overkill! For note that $a+b+c$ is even, and the parity of $x^n$ is the same as the parity of $x$. So $d$ is even. Since $1433$ is odd, $|d|\ne 2$, so $d$ cannot be prime.
Note that $|d|$ is also necessarily divisible by much smaller primes such as $3$ and $5$. This is perhaps an easier way to rule out $1433$. –  Erick Wong Jun 22 '12 at 21:02