Establish that ## a\equiv b\pmod {p} ##

[Math100]

Homework Statement

Assuming that ## a ## and ## b ## are integers not divisible by the prime ## p ##, establish the following:
If ## a^{p}\equiv b^{p}\pmod {p} ##, then ## a\equiv b\pmod {p} ##.

Relevant Equations

None.

Proof:

Suppose ## a^{p}\equiv b^{p}\pmod {p} ##, where ## a ## and ## b ## are integers not divisible by the prime ## p ##.
Then ## p\nmid a ## and ## p\nmid b ##.
Applying the Fermat's theorem produces:
## a^{p}\equiv a\pmod {p}, b^{p}\equiv b\pmod {p} ##.
Thus ## a^{p}\equiv a\equiv b^{p}\equiv b\pmod {p} ##.
Therefore, if ## a^{p}\equiv b^{p}\pmod {p} ##, then ## a\equiv b\pmod {p} ##.

 

[fresh_42]

Homework Statement:: Assuming that ## a ## and ## b ## are integers not divisible by the prime ## p ##, establish the following:
If ## a^{p}\equiv b^{p}\pmod {p} ##, then ## a\equiv b\pmod {p} ##.
Relevant Equations:: None.

Proof:

Suppose ## a^{p}\equiv b^{p}\pmod {p} ##, where ## a ## and ## b ## are integers not divisible by the prime ## p ##.
Then ## p\nmid a ## and ## p\nmid b ##.
Applying the Fermat's theorem produces:
## a^{p}\equiv a\pmod {p}, b^{p}\equiv b\pmod {p} ##.
Thus ## a^{p}\equiv a\equiv b^{p}\equiv b\pmod {p} ##.
Therefore, if ## a^{p}\equiv b^{p}\pmod {p} ##, then ## a\equiv b\pmod {p} ##.

Right, but what do you need ## p\nmid a ## and ## p\nmid b ## for?

If ##p\,|\,a^p\equiv b^p## then ##p\,|\,a ## and ##p\,|\,b## so ##a\equiv b\equiv 0\pmod{p}.##

 

[Math100]

Right, but what do you need ## p\nmid a ## and ## p\nmid b ## for?

If ##p\,|\,a^p\equiv b^p## then ##p\,|\,a ## and ##p\,|\,b## so ##a\equiv b\equiv 0\pmod{p}.##

I just realized that I don't need them for anything.

 

[Math100]

Right, but what do you need ## p\nmid a ## and ## p\nmid b ## for?

If ##p\,|\,a^p\equiv b^p## then ##p\,|\,a ## and ##p\,|\,b## so ##a\equiv b\equiv 0\pmod{p}.##

So should I include/add the line of ## p\,|\,a^p\equiv b^p##?

 

[WWGD]

I think you can argue that $$(a-b)$$ is always a factor of $$a^p-b^p$$and use that Integers are an integral domain.

 

[fresh_42]

I think you can argue that $$(a-b)$$ is always a factor of $$a^p-b^p$$and use that Integers are an integral domain.

Not if ##a\equiv b\pmod{p}.## Then you cannot divide.

 

[WWGD]

I just realized that I don't need them for anything.

Not if ##a\equiv b\pmod{p}.## Then you cannot divide.

I don't mean dividing. I mean, in an Integral domain, as in our case, if a.b=0, then either a=0 or b=0 ( mod p).

 

[Office_Shredder]

Sorry @fresh_42 I have always believed that 0 is a factor of 0.

I have never seen someone say ##x^2-1## factors into ##(x-1)(x+1)##, except when ##x=\pm 1##

 

[fresh_42]

Sorry @fresh_42 I have always believed that 0 is a factor of 0.

I have never seen someone say ##x^2-1## factors into ##(x-1)(x+1)##, except when ##x=\pm 1##

Yes, I thought he meant ##\dfrac{a^p-b^p}{a-b}=a^{p-1}+\ldots \pmod{p}##