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

$$\mathrm{Let}p\mathrm{be}\mathrm{a}\mathrm{prime}\mathrm{number}>3.\mathrm{What}$$$$\mathrm{is}\mathrm{the}\mathrm{remainder}\mathrm{when}{p}^2\mathrm{is}\mathrm{divided}\mathrm{by}$$$$12?$$

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

$$\mathrm{Any}\mathrm{prime}\mathrm{number}\mathrm{greater}\mathrm{than}$$$$3\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}$$$$6n+1$$$$6n-1(\mathrm{or}6\mathrm{n}+5)$$$$\mathrm{since}6n+2,6n+3,6n+4\mathrm{are}\mathrm{composites}.$$$${(6\mathrm{n}+1)}^2={36\mathrm{n}}^2+12\mathrm{n}+1$$$${(6\mathrm{n}+1)}^2\mathrm{mod}12=1$$$${(6\mathrm{n}-1)}^2={36\mathrm{n}}^2-12\mathrm{n}+1$$$${(6\mathrm{n}-1)}^2\mathrm{mod}12=1$$$$\mathrm{so}\mathrm{for}\mathrm{all}\mathrm{primes}p\mathrm{greater}>3$$$$p^2\mathrm{mod}12=1$$

Commented by RasheedSindhi last updated on 20/May/17

$${\mathit{e}}^{\mathit{x}}cellent!$$

Commented by mrW1 last updated on 20/May/17

$$Canyoupleaseexplainwhyanyprime$$$$isoftheform6n\pm 1?$$

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

$$allintegercanbewrittenas$$$$6n+mwhenm\in \{1,2,3,4,5\}$$$$6n+2=2(3n+1)notaprime$$$$6n+3=3(2n+1)notaprime$$$$6n+4=2(3n+2)notaprime$$$$\mathrm{so}\mathrm{only}\mathrm{two}\mathrm{possibilities}\mathrm{of}$$$$\mathrm{a}\mathrm{prime}\mathrm{number}>5,6n+1,6n+5$$$$wewrite6n+5as6n-1\mathrm{so}\mathrm{that}$$$$\mathrm{we}\mathrm{cover}\mathrm{number}5\mathrm{as}\mathrm{well}.$$$$6n+5=6(n+1)-1$$$$6n+1=6n-1$$$$\mathrm{so}\mathrm{all}\mathrm{primes}>3\mathrm{are}\mathrm{of}\mathrm{the}\mathrm{form}$$$$6n\pm 1$$

Commented by mrW1 last updated on 20/May/17

$$Thanksverymuch!$$

Commented by Tinkutara last updated on 21/May/17

$$\mathrm{It}\mathrm{is}\mathrm{also}\mathrm{called}{\mathrm{Euclid}}^{\prime }\mathrm{s}\mathrm{Division}\mathrm{Lemma}$$$$\mathrm{where}a=bq+r,0⩽r<b.$$

Commented by mrW1 last updated on 22/May/17

$$Accordingtothemethodof\mathit{P}\mathit{r}\mathit{a}\mathit{k}\mathit{a}\mathit{s}\mathit{h}$$$$wecanprove$$$$p^2\left(mod4\right)=1$$$$p^2\left(mod8\right)=1$$$$p^2\left(mod12\right)=1$$$$$$$$Thequestionisnowifthereexistsa$$$$furtherpositiveintegern>12sothat$$$$p^2\left(modn\right)=1foreveryprimenumberp$$$$uponacertainone?$$

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