Let-p-be-a-prime-number-gt-3-What-is-the-remainder-when-p-2-is-divided-by-12-

Question Number 13529 by Tinkutara last updated on 20/May/17

Let p be a prime number > 3 . What

is the remainder when p^{2} is divided by

12 ?

Commented by prakash jain last updated on 20/May/17

Any prime number greater than

3 is of the form

6 n + 1

6 n - 1 (or 6 n + 5)

since 6 n + 2, 6 n + 3, 6 n + 4 are composites .

{(6 n + 1)}^{2} = {36 n}^{2} + 12 n + 1

{(6 n + 1)}^{2} \mod 12 = 1

{(6 n - 1)}^{2} = {36 n}^{2} - 12 n + 1

{(6 n - 1)}^{2} \mod 12 = 1

so for all primes p greater > 3

p^{2} \mod 12 = 1

Commented by RasheedSindhi last updated on 20/May/17

e^{x} c e l l e n t!

Commented by mrW1 last updated on 20/May/17

C a n y o u p l e a s e e x p l a i n w h y a n y p r i m e

i s o f t h e f o r m 6 n \pm 1 ?

Commented by prakash jain last updated on 20/May/17

a l l i n t e g e r c a n b e w r i t t e n a s

6 n + m w h e n m \in {1, 2, 3, 4, 5}

6 n + 2 = 2 (3 n + 1) n o t a p r i m e

6 n + 3 = 3 (2 n + 1) n o t a p r i m e

6 n + 4 = 2 (3 n + 2) n o t a p r i m e

so only two possibilities of

a prime number > 5, 6 n + 1, 6 n + 5

w e w r i t e 6 n + 5 a s 6 n - 1 so that

we cover number 5 as well .

6 n + 5 = 6 (n + 1) - 1

6 n + 1 = 6 n - 1

so all primes > 3 are of the form

6 n \pm 1

Commented by mrW1 last updated on 20/May/17

T h a n k s v e r y m u c h!

Commented by Tinkutara last updated on 21/May/17

It is also called {Euclid}^{'} s Division Lemma

where a = b q + r, 0 ⩽ r < b .

Commented by mrW1 last updated on 22/May/17

A c c o r d i n g t o t h e m e t h o d o f P r a k a s h

w e c a n p r o v e

p^{2} (m o d 4) = 1

p^{2} (m o d 8) = 1

p^{2} (m o d 12) = 1

T h e q u e s t i o n i s n o w i f t h e r e e x i s t s a

f u r t h e r p o s i t i v e i n t e g e r n > 12 s o t h a t

p^{2} (m o d n) = 1 f o r e v e r y p r i m e n u m b e r p

u p o n a c e r t a i n o n e ?

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