Theorems on Modular Arithmetic

Question

Let $$a^x \equiv b^x \pmod p$$ and $$a^y \equiv b^y \pmod p$$ and $\gcd(y, x)=1$ then prove that $$a \equiv b \pmod p$$ As it is given that gcd$(y, x)=1$ so by the fundamental theorem of arithmetic there exists integers $u$ and $v$ such that $ux+vy=1$. Now without the loss of generality let us assume that $a < b$ then $b=a+k$ for some positive integer $k$. Now $$a+k=(a+k)^{ux+vy} \equiv a^{ux+vy}=a \pmod p.$$ So we get $p$ divides $k=b-a$. So $b \equiv a \pmod p$

$$$$$ Is My Proof Correct???

Welcome to Mathematics Stack Exchange. Did you mean $a\equiv b\bmod p$ where you wrote $x\equiv y\bmod p$ ? — J. W. Tanner, Feb 16 '20 at 17:52
You need to explain on $(a+k)^{ux+vy} \equiv a^{ux+vy}$, otherwise you are assuming what you are to prove. But I see no other problem — Aven Desta, Feb 16 '20 at 18:06
And for that. $(a+k)^{ux+vy} \equiv b^{ux+vy} \equiv (b^x)^u (b^y)^v \equiv (a^x)^u (a^y)^v \equiv a^{ux+vy}$ — Aven Desta, Feb 16 '20 at 18:10
Introducing the $k$ serves no apparent purpose. You could just as easily have written $b\equiv b^{ux+vy}\equiv a^{ux+vy}\equiv a \pmod p$. — lulu, Feb 16 '20 at 18:17

Aven Desta · Answer 1 · 2020-02-16T18:19:26.877

0

You need to explain on $(a+k)^{ux+vy} \equiv a^{ux+vy}$, otherwise you are assuming what you are to prove. But I see no other problem .

And for that. $(a+k)^{ux+vy} \equiv b^{ux+vy} \equiv (b^x)^u (b^y)^v \equiv (a^x)^u (a^y)^v \equiv a^{ux+vy}$

Edit: as @lulu said $k$ is not serving any purpose.

edited Feb 16 '20 at 18:19

answered Feb 16 '20 at 18:12

Aven Desta

610

But $u$ and $v$ both are not positive integers then how can we write $(b^x)^u(b^y)^v \equiv (a^x)^u(a^y)^v$ – user750786 Feb 17 '20 at 02:49
$a^u$ has integer solution, even when $u<0$ for modular arithmetic. $3^{-1} \equiv 5$ when working with modulo $7$. – Aven Desta Feb 17 '20 at 05:23
If that answered your question please mark it as solution. – Aven Desta Feb 17 '20 at 05:24

Theorems on Modular Arithmetic

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