Find the residue : \begin{equation} 1^{965}+2^{965}+...+2018^{965} \equiv x\ (\textrm{mod}\ 2021), x=? \end{equation} My attempt so far: \begin{equation} \varphi(965)=\varphi(5)\varphi(193)=768. \end{equation} But from Euler theorem applied for all numbers coprime with both 5 and 193 we have:\begin{equation} \Rightarrow\ 2^{768}\equiv 1\ (\textrm{mod}\ 2021) \Rightarrow\ 2^{965}\equiv 2^{197}\ (\textrm{mod}\ 2021). \end{equation} In the same fashion we get that \begin{equation} 3^{965}\equiv 3^{197}\ (\textrm{mod}\ 2021). \end{equation} Unfortunatelly this does not seem to work. Any ideas how to start?
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For Euler's theorem, you don't want the totient of the exponent, you want the totient of the modulus: $\varphi(2015)=1440$. So $a^{1440}\equiv 1$, for coprime $a$. Which, admittedly, doesn't help all that much. – Arthur Mar 08 '21 at 20:42
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Would it be easier if the sum went up to $2020^{965}$? – Hagen von Eitzen Mar 08 '21 at 20:48
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1@Arthur: did you mean $\phi(2021)=1932$ when you typed $\phi(2015)=1440?$ Your first clause was an important point for OP, though – J. W. Tanner Mar 08 '21 at 20:50
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1Suggest you look at Carmichael's reduced totient $\lambda(2021)=\text{lcm}(\lambda(43),\lambda(47)) = \text{lcm}(42,46) = 966$ – Joffan Mar 08 '21 at 20:50
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1@J.W.Tanner Yes. For some reason I had 2015 in my head. – Arthur Mar 08 '21 at 20:51
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Exploit reflection symmetry (modular negation) as explained in the linked dupe. – Bill Dubuque Mar 09 '21 at 09:38
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I think with solving mod equivalences, people are afraid of "dirty solutions". But I think just bashing it out works just fine.
Note that $$x^{2p+1}+(-x)^{2p+1}=(x+-x)(x^{2p})=0$$So, the sum is equal to $$1^{965}+2^{965}\bmod{2021}\to1+2^{965}\bmod{2021}$$Note that $2^{11}=2048$, so $2^{11}\equiv3^3\bmod{2021}$
Can you solve from here?
Rushabh Mehta
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Additionally, $2^{2\cdot 965+2}\equiv 1$, so we know that $(2(x-1))^2\equiv 1$, i.e., $2(x-1)$ can only by $\equiv \pm1$, which means $x-1\equiv \pm 1010$ – Hagen von Eitzen Mar 08 '21 at 20:58
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$(+1)$; or $2^{966}\equiv1\bmod2021$ by Euler, so $2^{965}\equiv 2^{-1}\equiv\dfrac{2021+1}2\bmod2021$ – J. W. Tanner Mar 08 '21 at 21:01
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@Don Thousand I am sorry but the first two lines are impossible for me to follow. The first line is trivial and I cannot see how did you use it. – Moscow1 Mar 09 '21 at 06:09
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