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Let $m$ be an arbitrary but fixed positive integer modulus, and let $x$, $y$, $z$, and $w$ be nonnegative integers. If $x \equiv y$ mod $m$, and $z \equiv w$ mod $m$, must it be the case that $x^z \equiv y^w$ mod $m$? I know that the corresponding statements for addition, subtraction, or multiplication in place of exponentiation is true in modular arithmetic. If the answer is no, I would like a counterexample with $m$ as small as possible. If there is a counterexample, what is the smallest possible modulus that admits a counterexample?

user107952
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    What have you tried? – jjagmath Jun 13 '25 at 01:40
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    What's the smallest $m$ that one could even try? – Eric Towers Jun 13 '25 at 01:42
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    Do you actually try to test this? At all? Try it with $\mod 5$. Not $2^7=128\equiv 3\pmod 5$ but $2^2 \equiv 4\pmod 5$. It fails. – fleablood Jun 13 '25 at 01:45
  • Smallest counter example is $m=3$ and $x\equiv y \equiv 2 \pmod 3$. Then $x^z\equiv 1\iff z$ is even (REGARDLESS as to what $z$ is equiv to $\mod 3$) and $x^w\equiv 2\iff w$ is odd. So $x^1\equiv 2 \not \equiv 1 \equiv y^4 \pmod 3$ is a counter example. – fleablood Jun 13 '25 at 01:50
  • @EricTowers the smallest one coult try would be $m=1$ in which case it is true because everything is $\equiv 0 \pmod 1$. Second smallest would be $2$ in which case it is true because $a^k \equiv a\pmod 2$ for any possible value of $k$. But it fails for any $m \ge 2$. – fleablood Jun 13 '25 at 01:53
  • @fleablood : Agreed. This is exactly the line of reasoning that OP should be able to understand and replicate. – Eric Towers Jun 13 '25 at 01:54
  • Cf. this answer – J. W. Tanner Jun 13 '25 at 02:00
  • Dang. Pity the question is closed. I was working on a full answer. As there are only $m$ possible values $\mod m$ there will have to be cases where $x^w\equiv x^z$ but $w\not = z$. But that does not have to do with $w\equiv z\pmod m$. And it may change for other values of $x$. There's a lot to when they do corelate but I can't get into details in a comment. But it is NOT via $w\equiv z \pmod m$ – fleablood Jun 13 '25 at 02:02
  • @fleablood Did you mean $m\ge 3$? – jjagmath Jun 13 '25 at 02:05
  • Oh, probably... where did I say that? Oh, there. Yes. It's true for $m=1$ because $K \equiv 0 \pmod 1$ for all $K$. And it is true for $m=2$ because $a^w \equiv a\pmod 2$ no matter what $w$ is. It is not true for $m\ge 3$. In particular if $p$ is prime then by Fermats little theorem $m^{p-1}\equiv 1\pmod p$ and in genera $a^{k+p}\equiv a^{k+1} \not \equiv a^k \pmod p$. But that's .... for later. – fleablood Jun 13 '25 at 02:10
  • The thing is when we add, subtract, multiply and even divide (which is also a troublesome trap) we are doing arithmetic on our input numbers. Exponents we are not doing any such arithmetic on the exponent power at all. All the exponent power is telling us is how many times we do something. It is not that we actually do anything with the number of the number of times itself except count them. – fleablood Jun 13 '25 at 02:18

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