Verify that ## x\equiv a^{p-2}b\pmod {p} ## is a solution

[Math100]

Homework Statement

Let $p$ be a prime and $gcd(a, p)=1$. Use Fermat's theorem to verify that $x\equiv a^{p-2}b\pmod {p}$ is a solution of the linear congruence $ax\equiv b\pmod {p}$.

Relevant Equations

None.

Proof:

Suppose $gcd(a, p)=1$ where $p$ is a prime.
Consider the linear congruence $ax\equiv b\pmod {p}$.
By Fermat's theorem, we have that $a^{p-1}\equiv 1\pmod {p}$.
Observe that $a^{p-1}\equiv 1\pmod {p}\implies a^{p-1}b\equiv b\pmod {p}\implies a(a^{p-2}b)\equiv b\pmod {p}$.
Thus $ax\equiv b\pmod {p}\implies x\equiv a^{p-2}b\pmod {p}$.
Therefore, $x\equiv a^{p-2}b\pmod {p}$ is a solution of the linear congruence $ax\equiv b\pmod {p}$.

 

[fresh_42]

Homework Statement:: Let $p$ be a prime and $gcd(a, p)=1$. Use Fermat's theorem to verify that $x\equiv a^{p-2}b\pmod {p}$ is a solution of the linear congruence $ax\equiv b\pmod {p}$.
Relevant Equations:: None.

Proof:

Suppose $gcd(a, p)=1$ where $p$ is a prime.
Consider the linear congruence $ax\equiv b\pmod {p}$.
By Fermat's theorem, we have that $a^{p-1}\equiv 1\pmod {p}$.
Observe that $a^{p-1}\equiv 1\pmod {p}\implies a^{p-1}b\equiv b\pmod {p}\implies a(a^{p-2}b)\equiv b\pmod {p}$.

Nice and correct so far.

Thus $ax\equiv b\pmod {p}\implies x\equiv a^{p-2}b\pmod {p}$.

This line is a bit confusing. You can simply remove this line.
You haven't solved for $x$. You have shown what you wrote next:

Therefore, $x\equiv a^{p-2}b\pmod {p}$ is a solution of the linear congruence $ax\equiv b\pmod {p}$.

$x$ set as $x:=a^{p-2}b$ solves the congruence, and this was all that had to be shown.
There are other numbers that solve it as well, namely all $x\in a^{p-2}b+p\mathbb{Z}.$

 

[Math100]

Observe that
\begin{align*}
&ax\equiv b\pmod {p}\implies ax\cdot a^{p-2}\equiv b\cdot a^{p-2}\pmod {p}\\
&\implies a^{p-1}x\equiv a^{p-2}b\pmod {p}.\\
\end{align*}
By Fermat's theorem, we have $a^{p-1}x\equiv x\pmod {p}$.
Thus $x\equiv xa^{p-1}\equiv a^{p-2}b\pmod {p}$.

 

[Math100]

Should I insert that above in my proof and delete what I wrote in my previous proof?

 

[fresh_42]

Should I insert that above in my proof and delete what I wrote in my previous proof?

Your first proof was perfect, just one line too many. You should remove the line before last. And if you want to be perfect then write:

...
Observe that $a^{p-1}\equiv 1\pmod {p}\implies a^{p-1}b\equiv b\pmod {p}\implies a(a^{p-2}b)\equiv b\pmod {p}$.

Therefore, $x:= a^{p-2}b\pmod {p}$ is a solution of the linear congruence $ax\equiv b\pmod {p}$.

 

[Math100]

Your first proof was perfect, just one line too many. You should remove the line before last. And if you want to be perfect then write:

...
Observe that $a^{p-1}\equiv 1\pmod {p}\implies a^{p-1}b\equiv b\pmod {p}\implies a(a^{p-2}b)\equiv b\pmod {p}$.

Therefore, $x:= a^{p-2}b\pmod {p}$ is a solution of the linear congruence $ax\equiv b\pmod {p}$.

When you were in college, did you pursue perfection as well?

 

[fresh_42]

When you were in college, did you pursue perfection as well?

Not really. I was more a student than a studier. That changed after the first exams.