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If $\gcd(a,35)=1$, then show that $a^{12}\equiv1\pmod{35}$

I have tried this problem beginning with $a^6 \equiv 1 \pmod{7}$ and $a^4 \equiv 1 \pmod{5}$ (Fermat's Theorem) but couldn't get far enough. Please help.

Martin Sleziak
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Saurabh
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    Good start... So, $a^{12}=$____$\pmod{7}$ and $a^{12}=$____$\pmod{5}$. Can you fill in? – Did Jun 08 '12 at 08:43
  • @did oh yes !!! 1,1 and I got it .Thanks for the hint.How could I miss that :( – Saurabh Jun 08 '12 at 08:50
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    You are welcome. Surely you know that you may write your own solution, post it here as an answer, wait a little bit, and accept it. It is even recommanded... – Did Jun 08 '12 at 08:57

3 Answers3

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Since $\gcd(a,7) =\gcd(a,5) = 1$, from Fermat's theorem,
$$a^6\equiv 1\pmod7 \quad \text{ and } \quad a^4\equiv1\pmod5. $$ Hence, $$ a^{12}\equiv 1\pmod7 \quad \text{ and } \quad a^{12}\equiv1\pmod5. $$ This means that $$7\mid a^{12}-1 \quad\text{ and } \quad 5\mid a^{12}-1. $$ Since $\gcd(7,5)=1$, $$35\mid a^{12}-1, $$ that is, $$ a^{12}\equiv1\pmod{35}. $$

Did
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Saurabh
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One way to do this is to use Euler's totient theorem. You are familiar with Fermat's little theorem, and this is a generalized version. This approach is lengthier, but instructive, because it shows that the totient function doesn't always give you the minimal exponent. For more information, see here.

From the theorem, we know

$$a^{\phi(35)} \equiv_{35} 1$$

A simple computation shows that $\phi(35)=24$ (this can be shortened if you know more about the $\phi$ function).

So we have

$$a^{24}\equiv 1. $$

If $k^2\equiv_{35} 1$, we see $(k-1)(k+1)\equiv 0$, so $k\equiv 1$, $k\equiv -1$, or $k\equiv \pm 6$, by using the prime factorization of $35$ to rule out other values. Applying this to the above, we will be done once we rule out the cases

$$a^{12}\equiv -1, \qquad a^{12}\equiv \pm 6$$

If $a^{12} \equiv -1$, we would need $(a^6)^2$ to be $-1$ mod $35$. To show $-1$ is not a square mod $35$, it suffices to show it is not a square mod $7$ (if $a^2$ were $-1$ mod 35, the same $a$ mod 7 would produce $a^2\equiv -1$ mod 7). It is straightforward to see by checking all possibilities that $-1$ is not the square of any number when considered modulo 7.

If $a^{12} \equiv 6$, then again $a^{12} \equiv - 1$ mod 7, and we reach a similar contradiction.

If $a^{12} \equiv -6$, then $(a^{3})^4 \equiv -1$ mod 5, which is impossible since $a^4 \equiv 1$ mod 5 for all $a$ by Euler's totient theorem.

Potato
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  • Interesting solution by residues, instead of the further extension of the original approach. Thanks for sharing. – awllower Jun 08 '12 at 12:20
  • @Potato In the $3\text{rd}$ paragraph how you had written "In general, if $k^2 \equiv 1$ we see....." How other cases are ruled out e.g. $k^2 \equiv 1 \pmod{3}$ then $k \equiv 1 \text{ or } k\equiv -1 \text{ or } k\equiv 4....$ – Saurabh Jun 20 '12 at 07:17
  • We have the factorization $(k-1)(k+1)\equiv_{35} 0$. The numbers on the left have to be 0 or divisors of $35$. I actually forgot to explicitly write out the case $k=6$, but that's not hard to see using the same methods I used. But $k\equiv 4$ won't work, for example, because $3*15=15$, which is not a multiple of $35$. – Potato Jun 20 '12 at 08:00
  • @Potato:If $\operatorname{gcd}(m,n) =1$ and $a \equiv c (\text{mod m})$, $b \equiv d (\text{mod n})$, then $ab \equiv cd (\text{mod mn})$. – Styles Feb 28 '18 at 07:00
  • $k\equiv-6$ is another possibility for $k^2\equiv1\pmod{35}$ – J. W. Tanner May 16 '24 at 02:56
  • @J.W.Tanner Thanks. I think it's right now. – Potato May 16 '24 at 03:04
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Note that:

  • If $a \equiv b \ (\text{mod m})$ then $a^{2} \equiv b^{2} \ (\text{mod m})$.

  • If $\operatorname{gcd}(m,n) =1$ and $a \equiv 1 (\text{mod m})$, $b \equiv 1 (\text{mod n})$, then $ab \equiv 1 (\text{mod mn})$.