<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] 
$\gcd(80,8a^2+1)$

Factors of $80$ are $8$ and $10$.So,if $\gcd\neq0$ then $8a^2+1$ must have $8$(or $2$ if you consider prime factors) or $10$ as factor.But clearly $8a^2+1$ cannot have $8$(or $2$) or $10$ as factors.Hence they have no common factors.hence,$\gcd=1$
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