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So I want to explain how one would go about solving the equation

$$y^{2019}\equiv 3571700849900719233 \quad (\text{mod} \ p),$$

where $p=2^{64}+13,$ which is the first prime after $2^{64}.$

I asked this same question a few days ago here, but the answer does not adequtely elaborate for example how one would know that $2019$ is relatively prime to $p-1?$ (without using software that is) And why does $2019\; \alpha \equiv 1 \bmod (p-1) \implies y=3571700849900719233^\alpha \bmod p?$

Is there a way to algorithmically simplify the equation?

Parseval
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    "how one would know that $gcd(2019,p-1)=1$?" This is easy. $2019=3\cdot 673$. Of course, $3$ does not divide $p-1=2^{64}+12$, since $3\nmid 2$. Similarly, with $673$. I think, Robert Isreal's answer is really clear. – Dietrich Burde Feb 18 '19 at 20:27
  • As for the second question, it comes from raising both sides of the equation to the $\alpha$ power and applying Fermat's little theorem – nkm Feb 18 '19 at 20:33
  • I think it's unreasonable to expect to handle questions involving $3571700849900719233$ or $2^{64}+13$ without some computation. $2^{64}+12$ is reasonably small as these things go...easy enough to check divisibly certainly. – lulu Feb 18 '19 at 20:34
  • The answer to the nontrivial part (2nd question) is already given in the answers in the dupe link on your prior question (e.g. see my answer). Did you read them? If you did not understand them then you should ask questions there (in comments) before posting further (duplicate) questions. – Bill Dubuque Feb 18 '19 at 20:46
  • @NathanielMayer To apply Ferma's don't I need the residual to be one? If $b=$ the big number, then doing what you say I get $$y^{2019\alpha}\equiv b^\alpha.$$ How is it applied here? – Parseval Feb 18 '19 at 20:48
  • @BillDubuque - Yeah I truely apologize for the duplicate, but if you look at the comments in that thread, I did read, and ask a followup question but there was no answer. So my only hope was to wait. – Parseval Feb 18 '19 at 20:50
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    But you didn't do as I suggest, i.e. ask questions in comments here if the answers there are not clear. There is no sign that you read that thread (which answers your 2nd question). If everyone posted a new thread everytime an answer needed elaboration the site would be a big mess. Please first pose questions in comments. If that fails then post another question. – Bill Dubuque Feb 18 '19 at 20:54
  • @BillDubuque - You're right, I totally missed this thread before. I'll have a look there. Is it ok if I delete this? Thanks for your feedback on this! – Parseval Feb 18 '19 at 20:58
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    @Parseval Fermat's little theorem says that $a^{p-1} \equiv 1\mod p$. Consequently if $n \equiv 1 \mod p-1$, then $a^n \equiv a \mod p$. – nkm Feb 18 '19 at 20:59
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    @Parseval That's fine, since there are no answers here (and it may save someone from posting another dupe answer). You can always pose another question if those answers (and elaboration) doesn't help. – Bill Dubuque Feb 18 '19 at 21:01
  • @Parseval How did you miss it? That's the thread listed as the dupe target on your prior question. – Bill Dubuque Feb 18 '19 at 21:05
  • @BillDubuque - Small screen, bad resolution and fast scrolling down to comments. That yellow bar statins its a dupe just flies up. Will be more careful next time and scroll slower. – Parseval Feb 18 '19 at 21:07

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