Show that $\gcd(80,8a^2+1)=1$

Question

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??

Answer 1

Squares are $\equiv 0,1,4\pmod 5$, hence $8a^2+1\equiv 1,9,33\equiv1,4,3\not\equiv 0\pmod 5$

Answer 2

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 $5$.

Answer 3

The problem easily reduces to $(8a^2+1,5)=1$

As $5$ is prime, either $5\mid(8a^2+1)$ or $(8a^2+1,5)=1$

$$8a^2+1\equiv0\pmod5\iff8a^2\equiv-1\equiv4\iff2a^2\equiv1\iff a^2\equiv3$$

Now $a\equiv0,\pm1,\pm2\pmod5\implies a^2\equiv0,1,4\not\equiv3\pmod5$

Answer 4

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

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,$\gcd=1$