$a^{12} \equiv 1 \pmod{35}$,knowing that $(a,35)=1$

Question

Prove that $\forall a \text{ with } (a,35)=1:$

$$a^{12} \equiv 1 \pmod{35}$$

$$35 \mid a^{12}-1 \Leftrightarrow 5 \cdot 7 \mid a^{12}-1 \overset{(5,7=1)}{ \Leftrightarrow} 5 \mid a^{12}-1 \text{ and } 7 \mid a^{12}-1$$

Therefore, $\displaystyle{ a^{12} \equiv 1 \pmod{35} \Leftrightarrow a^{12} \equiv 1 \pmod 5, a^{12} \equiv 1 \pmod 7}$

$$(a,35=1) \Rightarrow (a, 5 \cdot 7)=1 \overset{(5,7)=1}{\Rightarrow } (a,5)=1 \text{ and } (a,7)=1$$

According to Fermat's theorem:

$$a^4 \equiv 1 \pmod 5$$ $$a^{12} \equiv (a^4)^3 \equiv 1 \pmod 5$$

Also:

$$a^6 \equiv 1 \pmod 7$$ $$a^{12} \equiv (a^6)^2 \equiv 1 \pmod 7$$

So,we conclude that:

$$a^{12} \equiv 1 \pmod{35}$$

Could you tell me if it is right?

See http://mathworld.wolfram.com/CarmichaelFunction.html – lab bhattacharjee Jun 21 '14 at 15:53 — lab bhattacharjee, Jun 21 '14 at 15:53

score 1 · Accepted Answer · answered Jun 21 '14 at 15:59

1

Yes we can tell. This is because we have $a^{12} \equiv 1 \pmod 5 \iff 5|(a^{12}-1).$ Similarly $a^{12} \equiv 1 \pmod 7 \iff 7|(a^{12}-1)$. Since ${\rm gcd}(5,7)=1$, we have $5\cdot 7 | (a^{12}-1)$, i.e., $a^{12} \equiv 1 \pmod {35}$.

answered Jun 21 '14 at 15:59

Indrayudh Roy

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Nice,thank you very much!!!! – evinda Jun 21 '14 at 16:07
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The fact ${\rm gcd}(5,7)=1$ is essential here. It would not necessary hold if the gcd is more than $1$. – Indrayudh Roy Jun 21 '14 at 16:10
Nice,I understand..If I would like to prove that if $m \mid a, n \mid a \text{ and } (m,n)=1$,then $m \cdot n \mid a $,how could I do this? – evinda Jun 21 '14 at 16:12
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Let $a=mr$ and then show that $n | r$. – Indrayudh Roy Jun 21 '14 at 16:14
A ok,I got it...thank you very much!!!! – evinda Jun 21 '14 at 16:15
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You're welcome. :) – Indrayudh Roy Jun 21 '14 at 16:17

score 0 · Answer 2 · answered Jun 21 '14 at 16:49

Hint : For $(a,5)=1$ only possibilities are : $a=5k+1;5k+2;5k+3;5k+4$

$(5k+1)^{12}\equiv 1\mod5$ case should be clear...

$(5k+2)^{12}\equiv 1\mod 5$ should be clear if you know what $2^{12}$ is...

$(5k+3)^{12}\equiv 1\mod 5$ should be clear if you know what $3^{12}$ is...

$(5k+4)^{12}\equiv 1\mod 5$ should be clear if you know what $4^{12}$ is...

You may repeat this kind of strategy for case of $7$ and conclude what you wanted to...

score 0 · Answer 3 · answered Jun 21 '14 at 18:10

Hint $\ $ If $\,a\,$ is coprime to primes $\,p_1\!\ne p_2\,$ and $\,\color{#0a0}{p_i\!-1\mid n}\,$ then by $\rm\color{#c00}{little\ Fermat}$

$\qquad\qquad\qquad {\rm mod}\ p_i\!:\,\ a^n\equiv (\color{#c00}{a^{\,\large p_i-1}})^{\Large\color{#0a0}{\frac{n}{p_i-1}}\!}\equiv \color{#c00}1^{\Large\color{#0a0}{\frac{n}{p_i-1}}}\equiv 1\,\Rightarrow\, p_i\mid a^n-1\,\Rightarrow\, p_1 p_2\mid a^n-1$

$a^{12} \equiv 1 \pmod{35}$,knowing that $(a,35)=1$

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