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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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Hint Consider separately the moduli $3$ and $4$. This reduces significantly the number of cases to be chacked. – awllower May 23 '13 at 16:59
up vote 3 down vote accepted

$ 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: $\,3 < p\,$ prime $\,\Rightarrow\, p = 6n\pm 1\,\Rightarrow\, p^2 = 36n^2\pm12n+1$

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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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I think I get what you are saying.Thanks. – rahul May 23 '13 at 17:02

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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@Rahul, how about this one? – lab bhattacharjee May 23 '13 at 17:24
bhattacharjee ,hmmm I never expected this many answers,but I thuik it is a fairly good one. – rahul May 23 '13 at 17:30
@rahul, thanks. Hope I make my point clear. You may have a look into… – lab bhattacharjee May 23 '13 at 17:35


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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