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."