Compute $\gcd(83^9+1,83^9+83^2+1)$.
I notice that both numbers have the form $83^9+a$, where $a$ is a constant. So I can use the Euclidean algorithm to find their greatest common divisor (gcd). The Euclidean algorithm works by repeatedly applying the division algorithm, which says that for any integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a=bq+r$ and $0\leq r < b$. So I start by applying the division algorithm to the two numbers, where I take $a=83^9+83^2+1$ and $b=83^9+1$. I want to find $q$ and $r$ such that $83^9+83^2+1=(83^9+1)q+r$. To do this, I can try to estimate $q$ by dividing the leading terms of both sides. That is, I divide $83^9$ by $83^9$ and get $q=1$. This means that $r$ is the difference between $83^9+83^2+1$ and $83^9+1$. That is, $r=83^2$. So I have $83^9+83^2+1=(83^9+1)+83^2$. Now I repeat the division algorithm, but this time I take $a=83^9+1$ and $b=83^2$. I want to find $q$ and $r$ such that $83^9+1=83^2q+r$. Again, I can try to estimate $q$ by dividing the leading terms of both sides. That is, I divide $83^9$ by $83^2$ and get $q=83^7$. This means that $r$ is the difference between $83^9+1$ and $83^2\cdot 83^7$. That is, $r=1$. So I have $83^9+1=83^2\cdot 83^7+1$. Now I have found the gcd, because the division algorithm says that the gcd is the last nonzero remainder in the process. That is, $\gcd(83^9+1,83^9+83^2+1)=1$. 