<s>[INST] Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ Find the greatest common divisor of $2003^4 + 1$ and $2003^3 + 1$ without the use of a calculator. It is clear that $2003^4+1$ has a $082$ at the end of its number so $2003^4+1$ only has one factor of 2, while $2003^3+1$ has a $028$ at the end of its number so  $2003^3+1$ has 2 factors of 2. The answer is supposed to be 2, but at this point it could be greater than 2. Any ideas? This was on a previous qualifying exam at BU.
 [/INST] $d = \text{gcd}(2003^4+1,2003^3+1) \Rightarrow d \mid 2003^4-2003^3 = 2003^3\cdot 2002 = 2003^3\cdot 2\cdot 7\cdot 11\cdot 13$, and by checking case by case, we have $2$ is the only number such that $d$ divides into evenly. So $d = 2$.
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