Show that ## 168=3\cdot 7\cdot 8 ## divides ## a^{6}-1 ##

[Math100]

Homework Statement

If $gcd(a, 42)=1$, show that $168=3\cdot 7\cdot 8$ divides $a^{6}-1$.

Relevant Equations

None.

Proof:

Suppose $gcd(a, 42)=1$.
Then $42=2\cdot 3\cdot 7$ and $gcd(a, 2)=gcd(a, 3)=gcd(a, 7)=1$.
Applying the Fermat's theorem produces:
$a\equiv 1\pmod {2}, a^{2}\equiv 1\pmod {3}$ and $a^{6}\equiv 1\pmod {7}$.
This means $a^{6}=(a^{2})^{3}\equiv 1\pmod {3}$.
Observe that $a^{6}-1=(a^{3}-1)(a^{3}+1)=(a-1)(a+1)(a^{2}+a+1)(a^{2}-a+1)$.
Since $a$ is odd, it follows that $a>0\implies a\geq 3$.
Now we have $a^{2}\equiv 1\pmod {8}\implies (a^{2})^{3}\equiv 1^{3}\pmod {8}\implies a^{6}\equiv 1\pmod {8}$.
Thus, $168\mid (a^{6}-1)$ because $3, 7, 8$ are relatively prime to each other.
Therefore, if $gcd(a, 42)=1$, then $168=3\cdot 7\cdot 8$ divides $a^{6}-1$.

 

[fresh_42]

Homework Statement:: If $gcd(a, 42)=1$, show that $168=3\cdot 7\cdot 8$ divides $a^{6}-1$.
Relevant Equations:: None.

Proof:

Suppose $gcd(a, 42)=1$.
Then $42=2\cdot 3\cdot 7$ and $gcd(a, 2)=gcd(a, 3)=gcd(a, 7)=1$.
Applying the Fermat's theorem produces:
$a\equiv 1\pmod {2}, a^{2}\equiv 1\pmod {3}$ and $a^{6}\equiv 1\pmod {7}$.
This means $a^{6}=(a^{2})^{3}\equiv 1\pmod {3}$.
Observe that $a^{6}-1=(a^{3}-1)(a^{3}+1)=(a-1)(a+1)(a^{2}+a+1)(a^{2}-a+1)$.
Since $a$ is odd, it follows that $a>0\implies a\geq 3$.
Now we have $a^{2}\equiv 1\pmod {8}\implies (a^{2})^{3}\equiv 1^{3}\pmod {8}\implies a^{6}\equiv 1\pmod {8}$.
Thus, $168\mid (a^{6}-1)$ because $3, 7, 8$ are relatively prime to each other.
Therefore, if $gcd(a, 42)=1$, then $168=3\cdot 7\cdot 8$ divides $a^{6}-1$.

Right, but what did you use the factorization of $a^6-1$ for? Instead, you should have explained $a^2\equiv 1\pmod{8}.$ E.g. by $(2k+1)^2=4k(k+1)+1$ where either $k$ or $k+1$ is even.

 

[Math100]

Right, but what did you use the factorization of $a^6-1$ for? Instead, you should have explained $a^2\equiv 1\pmod{8}.$ E.g. by $(2k+1)^2=4k(k+1)+1$ where either $k$ or $k+1$ is even.

Absolutely nothing. Now, I just realized that I shouldn't factor out $a^{6}-1$, because like you mentioned, I didn't use this on anything in this proof. Also, for the $a^{2}\equiv 1\pmod {8}$ part, are you saying that letting $a=2k+1$ where $k=2m$ for some $k\in\mathbb{Z}$ can help better explain where we got the $a^{2}\equiv 1\pmod {8}$ part from?

 

[fresh_42]

Absolutely nothing. Now, I just realized that I shouldn't factor out $a^{6}-1$, because like you mentioned, I didn't use this on anything in this proof. Also, for the $a^{2}\equiv 1\pmod {8}$ part, are you saying that letting $a=2k+1$ where $k=2m$ for some $k\in\mathbb{Z}$ can help better explain where we got the $a^{2}\equiv 1\pmod {8}$ part from?

Number theorists have it probably in mind: $a\equiv 1\pmod{2}\Longrightarrow a^2\equiv 1\pmod{8}$ but I had to use this additional line. From $a\equiv 1\pmod{2}$ we get $a=2k+1$ and $a^2=4k(k+1)+1.$ This shows $a^2\equiv 1\pmod{8}$ no matter whether $k$ is odd or even because $\{k,k+1\}$ always contains an even number.

 

[Math100]

Number theorists have it probably in mind: $a\equiv 1\pmod{2}\Longrightarrow a^2\equiv 1\pmod{8}$ but I had to use this additional line. From $a\equiv 1\pmod{2}$ we get $a=2k+1$ and $a^2=4k(k+1)+1.$ This shows $a^2\equiv 1\pmod{8}$ no matter whether $k$ is odd or even because $\{k,k+1\}$ always contains an even number.

I have to use this additional line, too, given the fact that I am not a number theorist. In fact, I got the $a^{2}\equiv 1\pmod {8}$ part, mainly because $a$ is odd and so $a>0, a\geq 3$, that's why I plugged in $a=3$ and got $a^{2}=9, a^{2}\equiv 1\pmod {8}$, but I think this is vague, just as you mentioned above, I should have explained this. Thank you for the additional line.