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I know that $\forall{k}\in\mathbb{N}: a \equiv b \pmod m \implies a^k \equiv b^k \pmod m$.

But the converse is not always true.

A few simple counterexamples:

  • $2^2 \equiv 1^2 \pmod 3$
  • $2^3 \equiv 1^3 \pmod 7$

I am wondering if there are any theorems where the converse actually holds with some particular conditions being imposed on the variables.

I tried googling but hadn't found anything useful on it yet.

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    Have you look at Euler's theorem (https://en.wikipedia.org/wiki/Euler's_theorem)? – 2017-01-01
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    For $k$ even it's never true (except for $m = 1, 2$) because $a^k \equiv (-a)^k$. For odd $k$, it's a more interesting question. – 2017-01-01
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    I don't agree with the downvotes for this question. I think some people just downvote questions that aren't contrived problems from contests or textbooks. – 2017-01-01
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    @JosuéTonelli-Cueto Links may be presented in a [title here](link here) format. – 2017-01-01
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    @SimpleArt Thank you for the tip! – 2017-01-01

2 Answers 2

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If $a^n=b^n, \pmod{p}$ implies $a=b \pmod p$ then for this to be always true, then $\gcd(n,p-1)=1$. This means, for example, $2^5=x^5 \pmod 7$ implies that $x-2$ is a multiple of 7.

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    Thank you so much for this answer. It seems pretty useful. I tried proving it. In case anyone's interested in the proof: by [Fermat's Little Theorem](https://en.m.wikipedia.org/wiki/Fermat's_little_theorem), $a^{p-1} \equiv b^{p-1} \pmod p$. Combining this with the initial condition, we get, $a^{\gcd(n,p-1)} \equiv b^{\gcd(n,p-1)} \pmod p$. So if $\gcd(n,p-1) = 1$ then $a \equiv b \pmod p$. – 2017-01-01
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A sufficient condition is that the kernel of the groups-homomorpgism $$\phi _k (x)=x^k $$ is trivial. If it is not trivial you will always get counter examples.

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    Might be overkill for a question about elementary number theory. – 2017-01-01
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    This doesn't answer anything. When is the kernel trivial? – 2017-01-01
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    I beg the differ. The way to approach problems we modulo is exactlu to treat them as groups. – 2017-01-01
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    yoni, you didn't treat anything. You just rewrote the statement. If you solve the problem with group theory introduced, you will have given an answer. So far, it is not one. – 2017-01-01