Show that if $3\mid(a^2+1)$ then $3$ does not divide $(a+1)$. Show that if $3\mid(a^2+1)$ then $3$ does not divide $(a+1)$. 
using proof of contradiction 
can someone prove this using contradiction method please
 A: For the original question, if $3 \nmid a$, then $3 \nmid a^2$ since $3$ is prime (consider the factorization of $a$ and $a^2$), but this doesn't imply anything about whether $3$ divides $a^2 + 1$.
Here's a different approach: Suppose $3$ is a divisor of both $a^2 + 1$ and $a + 1$. Can you justify why $$3 \mid \Big( a(a + 1) - (a^2 + 1)\Big) = a$$
and why that's a contradiction?
A: $3\mid \color{blue}{a^2\!+\!1},\,\color{#0a0}{a\!+\!1}\,\Rightarrow\,3\mid\color{blue}{a^2\!+\!1}-\overbrace{(\color{#0a0}{a\!+\!1})(a\!-\!1)}^{\large a^2\,-\,1} = \color{#c00}2,\,$ a contradiction, completing your proof.
Or modly: $\ 3\mid a\!+\!1\,\Rightarrow\, {\rm mod}\ 3\!:\ a\equiv -1\,\Rightarrow\ a^2\!+\!1\equiv \color{#c00}2\,\Rightarrow\,3\nmid a^2\!+\!1$
Remark $\ $ The first proof shows more generally that $\,\gcd(a^2+1,a+1)\mid 2.\,$ Using parity  we see further that the gcd $= 2 \iff a$ is odd; otherwise the gcd $= 1\ \ (\!\iff a\,$ is even$)$.
A: Prove the contrapositive:

If $3\mid a+1$, then $3\nmid a^2+1$.

Which is easy, because then $a=3k-1$ and $a^2+1=9k^2-6k+2$, which isn't divisible by $3$.
A: Hint: just consider the 3 cases
$$a = 3k, a=3k+1, a=3k-1
$$
