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Nothing I throw at this will stick, Eulers theorem, Fermats theorem feels useless when I apply them on large numbers like this. I'm interested in solving $x^{a}\equiv b \ (\text{mod} \ p),$ where $a=2019$, $b=3571700849900719233$ and $p=2^{64}+13.$

This is what I've done so far:

$$x^{a}\equiv b \ (\text{mod} \ p)\Leftrightarrow x^{a}=b+np, \ \text{for some} \ n\in \mathbb{N}.$$

This means that $\text{gcd}(x,p)=1$ so

$$x^{\phi(p)}=x^{\phi(p-1)}=x^{2^{64}+12}\equiv 1 \ (\text{mod} \ 2^{64}+13),$$

which is exactly Fermats theorem. I also realise that $$x^{2^{64}+12}=x^{2^{64}}x^{2^3}x^{2^2}=(x^{2})^{69}\equiv 1^{69} \ (\text{mod} \ 2^{64}+13),$$

which is equivalent to

$$x^{2}\equiv 1 \ (\text{mod} \ 2^{64}+13).$$

Which feels very wrong and still very hard to proceed. How are these huge numbers treated?

Parseval
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1 Answers1

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Hint: $p = 2^{64}+13$ is prime, and $2019$ is coprime to $p-1$. If $2019\; \alpha \equiv 1 \bmod (p-1)$, then your answer will be $b^\alpha \bmod p$.

Robert Israel
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  • I don't really follow. What is $\alpha$? How does this relate to my $x?$ Also, $b^{\alpha} \equiv 14757161404939787331 \ \text{mod} \ p ,$ which is not the correct answer for $x$. – Parseval Feb 14 '19 at 16:01
  • If $\alpha$ is the multiplicative inverse of $2019$ mod $(p-1)$, then $x \equiv b^\alpha \bmod p$. It is not $14757161404939787331$. – Robert Israel Feb 14 '19 at 16:31
  • I don't understand how $2019\alpha \ \text{mod} (p-1)\Rightarrow x\equiv b^{\alpha}$. What theorems are you using? What reason are you using? I could use some more elaboration, however the answer you provided is infact correct. I just want to understand it properly. Also, according to maple, solving for $\alpha$ gives $\alpha=7830044413654822063$ so then I have to compute $$3571700849900719233^{7830044413654822063} \equiv \text{mod}(2^{64}+13).$$

    This seems really hard to compute by hand on an exam.

     – Parseval Feb 14 '19 at 17:48
  • @Parseval Do you have any reason to believe that it was designed to be computed completely by hand? – Bill Dubuque Feb 18 '19 at 20:43
  • Nothing about $3571700849900719233$ seems designed to be computed completely by hand. – Robert Israel Feb 19 '19 at 03:55