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Consider the congruence equation $$ 3^x \equiv -1~(\text{mod}~10).$$

What are the solutions ?

We can see $x=2$ is a solution and thus by cyclic property $2+10k,~k \in \mathbb Z$ should be a solution. But it is not. For $k=1$, $x=12$ is not a solution. Why is so ?

Instead, $x=6, ~10,~14, \cdots, 2+4k$ are solution.

Why is $x=2+10 k$ not a solution ?

MAS
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  • "by cyclic property... +10k..." You seem to be confusing and conflating a property you've heard about things which occur "on the floor" as things being added or multiplied and incorrectly thinking it also applies to variables occurring inside of exponents or other more arbitrary functions. – JMoravitz May 08 '23 at 16:48
  • "Why is... not a solution" You yourself have already shown why not... because it isn't. You yourself have already disproven this by noting that $x=2,6,10,14,\dots$ are the solutions here. – JMoravitz May 08 '23 at 16:49
  • @JMoravitz, I am still not getting into the deep part. We are taking solutions modulo $10$ i.e., in the ring $\mathbb Z_{10}$. So $2 \equiv 2+10k $ in $\mathbb Z_{10}$. – MAS May 08 '23 at 16:51
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    $!\bmod 10!:\ 3,$ has order $\color{#c00}4,$ so $,3^n\equiv 3^2\equiv -1\iff 3^{n-2}\equiv 1\iff \color{#c00}4\mid n-2\iff n = 2+4k,$ by the Order Theorem in the linked dupe. – Bill Dubuque May 08 '23 at 16:54
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    "We are taking solutions modulo..." That may be relevant for numbers who are added or multiplied to one another, however that is irrelevant to an exponent. Exponents are still meant to be interpreted in the same way... here $x^5 = x\times x\times x\times x\times x$ and the number $5$ is the usual number $5$ and it is not affected by other considerations based on what ring we are working in. – JMoravitz May 08 '23 at 16:56
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    @JMoravitz, Oh I got it. Thanks. We are dealing with exponents here. – MAS May 08 '23 at 16:57
  • @BillDubuque, why $3^n \equiv 3^2 \equiv -1 \Leftrightarrow 3^{n-2} \equiv 1$ mod $(10)$ ? – MAS May 11 '23 at 16:25
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    $,3^n\equiv -1\color{#0a0}{\Leftarrow}!\color{#c00}{\Rightarrow} 3^{n-2}\equiv 1$ follows by dividing $(\color{#c00}{\Rightarrow)}$ or multiplying $(\color{#0a0}{\Leftarrow})$ by $,3^2\equiv -1\ \ $ – Bill Dubuque May 11 '23 at 16:41
  • @BillDubuque, ahh nice. Thank you very much – MAS May 12 '23 at 05:06

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