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 $\dfrac{a^p-b^p}{a-b}=a^{p-1}+\ldots \pmod{p}$