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(The other similar post didn't help in this case)

I understand that $5^{119}\equiv 7(\mod59)$, but how can I show that the answer is indeed $7$, and not for example $(7 - 59)$ or $(7 + 59)$?

Thank you.

QuantumSuperfield
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jabben_perhaps
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    All three of those answers would be sufficient, as $a\equiv b \pmod c$ implies that $a\equiv b+c\times n \pmod c$ for any integer $n$. Sometimes people use modular notation to refer to the least positive remainder....if that's what you mean, then only $7$ is correct, since $7-59$ is negative and $7+59$ is greater than $7$. – lulu Dec 03 '23 at 12:02
  • It is not true that the answer "is indeed $7$". For this, you need to add more information (but not in the title, better in the post). – Dietrich Burde Dec 03 '23 at 12:16
  • The remainder $, r = a\bmod n,$ is the (unique) natural $,r < n,$ such that $,r\equiv a\pmod{!n},,$ i.e. the least natural $\equiv a\pmod{!n},$ see the linked dupes. Hence $,r-n<0,$ is too small (not natural), and $,r+n,$ is too large (not the least) to be the remainder. – Bill Dubuque Dec 03 '23 at 18:05
  • The calculation of $5^{119}\bmod 59$ by FLT and mod order reduction is already done here. – Bill Dubuque Dec 03 '23 at 18:16

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Do you know Euler's theorem, or at least that for any integer $\;a\;$ and any prime $\;p\;$ we have that $\;a^p\equiv a \mod p\;$ ? If you do then it is easy ( the following is done modulo $\;59\;$):

$$119=59\cdot 2 +1\implies 5^{119}=5^{2\cdot 59+1}=\left(5^{59}\right)^2\cdot 5=5^2\cdot 5=125=59\cdot 2+7=7$$

Of course, any other answer equals $\;7+59k\;,\;\;k\in\Bbb Z\;$ , so there are infinite answers to that problem.

DonAntonio
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    What do you mean infinite answers? The reminder of something divided by 59 has to be somewhere between $0,1,\cdots,58$ – Ariel Fishbein Dec 03 '23 at 13:20
  • @ArielFishbein I understand the problem given by the OP as a problem of equivalence classes mdulo $59$, that's all. It is not true that we must always work with the residues $;0,1,2,...,58\pmod{59};$, though that's what is almost always used in this kind of problems. Any other full list of representatives will work just fine. – DonAntonio Dec 03 '23 at 13:27
  • But you said that $5^{119}(mod 59)$ can be equal to more than one asnwer which is no true. How do you explain then – Ariel Fishbein Dec 03 '23 at 13:36
  • @ArielFishbein Of course it is true, as $$;7\pmod{59}=66\pmod{59}=-52\pmod{59}=125\pmod{59}=\ldots$$ – DonAntonio Dec 03 '23 at 15:40
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Dec 03 '23 at 18:17
  • I meant that you said that $x$ has infinite sulotions if $5^{119}=x(mod 59)$ – Ariel Fishbein Dec 03 '23 at 18:20