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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 $12$?

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    $\begingroup$ Hint Consider separately the moduli $3$ and $4$. This reduces significantly the number of cases to be chacked. $\endgroup$
    – awllower
    May 23, 2013 at 16:59

5 Answers 5

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$ 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.

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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.

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  • $\begingroup$ I think I get what you are saying.Thanks. $\endgroup$
    – rahul
    May 23, 2013 at 17:02
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Hint: $\,3 < p\,$ prime $\,\Rightarrow\, p = 6n\pm 1\,\Rightarrow\, p^2 = 36n^2\pm12n+1$

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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 $(3,8)=24$

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  • $\begingroup$ @Rahul, how about this one? $\endgroup$ May 23, 2013 at 17:24
  • $\begingroup$ bhattacharjee ,hmmm I never expected this many answers,but I thuik it is a fairly good one. $\endgroup$
    – rahul
    May 23, 2013 at 17:30
  • $\begingroup$ @rahul, thanks. Hope I make my point clear. You may have a look into math.stackexchange.com/questions/398656/… $\endgroup$ May 23, 2013 at 17:35
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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 ?

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