Show that ## 133\mid (a^{18}-b^{18}) ##.

[Math100]

Homework Statement

If $gcd(a, 133)=gcd(b, 133)=1$, show that $133\mid (a^{18}-b^{18})$.

Relevant Equations

None.

Proof:

Suppose $gcd(a, 133)=gcd(b, 133)=1$.
Then $133=7\cdot 19$.
Applying the Fermat's theorem produces:
$a^{6}\equiv 1\pmod {7}, b^{6}\equiv 1\pmod {7}$,
$a^{18}\equiv 1\pmod {19}$ and $b^{18}\equiv 1\pmod {19}$.
Observe that
\begin{align*}
&a^{6}\equiv 1\pmod {7}\implies (a^{6})^{3}\equiv 1^{3}\pmod {7}\implies a^{18}\equiv 1\pmod {7}\\
&b^{6}\equiv 1\pmod {7}\implies (b^{6})^{3}\equiv 1^{3}\pmod {7}\implies b^{18}\equiv 1\pmod {7}.\\
\end{align*}
This means $a^{18}-b^{18}\equiv (1-1)\pmod {7}\equiv 0\pmod {7}$ and $a^{18}-b^{18}\equiv (1-1)\pmod {19}\equiv 0\pmod {19}$.
Thus, $7\mid (a^{18}-b^{18})$ and $19\mid (a^{18}-b^{18})$.
Since $gcd(7, 19)=1$, it follows that $(7\cdot 19)\mid (a^{18}-b^{18})\implies 133\mid (a^{18}-b^{18})$.

 

[fresh_42]

Yes. Correct. Someone should trace these congruences, they are funny.

 

[Math100]

Yes. Correct. Someone should trace these congruences, they are funny.

How so?

 

[fresh_42]

How so?

I mean $133 \,|\, (a^{18}-b^{18})$ for all even numbers, or all multiples of $3$ and more?
Or $35\,|\,(a^{12}-1)$ for all even numbers, or all multiples of $3$ and more?

I find this funny. But I also found
$$2^n+7^n+8^n+18^n+19^n+24^n=3^n+4^n+12^n+14^n+22^n+23^n \text{ for } n =0,1,...,5$$
funny. Maybe my strange kind of humor. I remember that I once made a remark while walking back from a pizzeria and one friend laughed while the other one asked: "You understood this remark?" and the first one answered: "No, I have learned when to laugh."