Proving the remainder is $1$ if the square of a prime is divided by $12$ Given, $p$ is a prime number and $p>3$. How do we prove that the remainder $r$ is always $1$ if $p^2$ is divided by $12$?
 A: $ p^2 -1 =(p-1)(p+1) $. Since $p$ is odd, these are two even factors, so the product is divisible by $4$. Also, one of $p-1$, $p$, $p+1$ must be divisible by $3$, but since $p$ is prime, it is not divisible by $3$. Thus one of $p-1$, or $p+1$ must be divisible by three.
A: Hint: consider $0,1,2,\ldots, 11$, which are all the possible remainders modulo 12.  Take out the ones that couldn't be congruent to a prime bigger than 3.  Square all the ones that are left modulo 12, and see what you get.
A: Hint: $\,3 < p\,$ prime $\,\Rightarrow\, p = 6n\pm 1\,\Rightarrow\, p^2 = 36n^2\pm12n+1$
A: If $(a,12)=1, (a,3)=1$ and $ (a,2)=1$
$(a,3)=1\implies a\equiv\pm1\pmod 3\implies a^2\equiv1\pmod 3$
$(a,2)=1\implies a$ is odd $=2b+1$(say) where $b$ is some integer
$(2b+1)^2=4b^2+4b+1=8\frac{b(b+1)}2+1\equiv1\pmod 8$
$\implies a^2\equiv1\pmod { \text{lcm}(3,8)}$
Now, lcm $(3,8)=24$
A: Hint:
since $p>3$ we know that $p$ is not even.
So the posibilities for $p$ mod$(4)$ are :  $...$   
consequently for $p^2$ mod$(4)$ are : $...$
and for $p$ mod$(3)$ are : $...$  
consequently for $p^2$ mod$(3)$ are : $...$
If you did the right calculations you will see that in both cases $p^2\equiv 1$ mod$(4)$ and mod$(3)$ , so ?
