How Does Fermat's Little Theorem Apply to n^{39} \equiv n^3 (mod 13)?

[AryaSravaka]

Dear sir/madam
I have tried to do this question but can not figure it out. I gave up, but google gave me physicsforums site. I am very greatful and thanks for being genorous.

Thanks.
question is attached

 

[micromass]

What did you try already?? If you tell us where you're stuck, then we'll know how to help...

 

[AryaSravaka]

What did you try already?? If you tell us where you're stuck, then we'll know how to help...

Sir, I really can not see any connection between this problem and FLT ..Pls give me some insight how to start out .. got 1day left :)
thanks

 

[stringy]

We know $n^{13} \equiv n \ ( mod \ 13)$, right? That's the theorem.

But we also know that if $a' \equiv a \ ( mod \ m )$ and $b' \equiv b \ ( mod \ m )$, then $a'b' \equiv ab \ ( mod \ m )$. This implies

$$n^{39} \equiv n^3 \ (mod \ 13).$$

But then what's another way to express this congruence? To say that $n^{39}$ is congruent to $n^3$ means 13 divides what?

 

[AryaSravaka]

We know $n^{13} \equiv n \ ( mod \ 13)$, right? That's the theorem.

But we also know that if $a' \equiv a \ ( mod \ m )$ and $b' \equiv b \ ( mod \ m )$, then $a'b' \equiv ab \ ( mod \ m )$. This implies

$$n^{39} \equiv n^3 \ (mod \ 13).$$

But then what's another way to express this congruence? To say that $n^{39}$ is congruent to $n^3$ means 13 divides what?

Dear Sir, Thanks very much for your time. Anyway FLT was not in the exam.. I had to do 4 questions but i did 6 questions... Well I passed it..Yehiiiiiiiiiiiiiiiiiiiiii


I am truly greatful
with metta