<s>[INST] Show that $\gcd(80,8a^2+1)=1$ Show that $\gcd(80,8a^2+1)=1$
Let $\gcd(80,8a^2+1)=d$, then we have: $d|80a^2+10,80a^2\Rightarrow\ d|10$
So $d=1\ or\ 2\ or\ 5\ or\ 10$
Obviously $d$ can't be $2\ or\ 10$,but how can we show $d$ can't be 5??
 [/INST] Since $0^1 \equiv 0, 1^2 \equiv 1 \equiv 4^2, 2^2\equiv 4 \equiv 3^2 \mod 5$ we have $8a^2+1 \equiv 1$ or $4$ or $3 \mod 5$. This implies that $8a^2+1$ is not divisible by $5$.
</s>