Does congruence mod k work for exponents?

Asked Apr 07 '23 at 05:47

Active Apr 07 '23 at 08:06

Viewed 61 times

For $\text{mod 5}$,

$[1] = \{\dotsc, -4, 1, 6, \dotsc\}$

So, we can write,

$$6\equiv1\ \text{(mod 5)}\tag{1}$$

Which means that I can substitute $1$ for $6$ (and vice versa) anywhere I see it $(\text{mod 5})$. So, consider the example below:

$$6^{719} \ \text{mod 5}$$

$$=1^{719} \ \text{mod 5}$$

$$=1\ \text{mod 5}$$

$$=1$$

Now, consider another example,

$$2^6\ \text{mod 5}$$

$$=2^1\ \text{mod 5}\ \ \text{[Using (1)]}$$

$$=2$$

However, this is a wrong answer. The correct answer is 4. Why is substituting 1 for 6 not working in this case? Aren't they congruent to each other?

edited Apr 07 '23 at 08:06

Bill Dubuque

asked Apr 07 '23 at 05:47

tryingtobeastoic

Short answer: $6 \equiv 1 \pmod 5$ doesn't mean you can substitute $1$ for $6$ wherever you see it, so you're attempting to apply a rule that doesn't exist. – Brian Moehring Apr 07 '23 at 06:05
@BrianMoehring I see. So, what is the actual rule then? – tryingtobeastoic Apr 07 '23 at 06:10
Are you asking for the rule that applies in your example of $6^{719} \equiv 1 \pmod{5}$ or are you asking for the rule that applies to exponents? – Brian Moehring Apr 07 '23 at 06:18
@BrianMoehring Both please – tryingtobeastoic Apr 07 '23 at 06:24
1

See this answer for the first question. The second question is more involved, but the answer here covers the basic idea. In your particular example, Fermat's little theorem would be sufficient to show $$2^6 \equiv 2^{6\operatorname{mod} 4} \pmod{5}$$ – Brian Moehring Apr 07 '23 at 06:50
1

The correct rule for exponents is that you can reduce them mod any period $,\color{#c00}n,,$ i.e. $,a^{\large\color{#c00}n}\equiv 1,\Rightarrow, a^{\large k}\equiv a^{\large k\bmod{\color{#c00}n}},,$ see mod order reduction in the first linked dupe. See the 2nd linked dupe for sums and products (i.e. polynomial expressions). – Bill Dubuque Apr 07 '23 at 08:01

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