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I am trying to evaluate $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$. I have found that $15^{15} \equiv 1 \pmod{7}$ and that $16^{16} \equiv 2 \pmod{7}$.

To evaluate $15^{15} \pmod{17}$, I did the following: $$15 = 2 \times 7 + 1 \equiv 1 \pmod{7}$$ $$15^{15} \equiv 1 \pmod{7}$$

Then, to evaluate $16^{16}$, I wrote: $$16 = 15 + 1 \equiv 1 + 1 = 2 \pmod{7}$$ $$16^{16} \equiv 2^{16} \pmod{7}$$

$$2^{3} = 8 = 7+1 \equiv 1 \pmod{7}$$ $$2^{16} = 2^{3} \times 2^{13} \equiv 2^{13} = 2^{3} \times 2^{10} \equiv 2^{10} \equiv \dots \equiv 2 \pmod{7}$$

Nevertheless, I have not managed to figure out how to evaluate $17^{17}$. How should I go about this and is my overall approach for evaluating the sum in question a good one?

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  • $\begingroup$ Use Fermat's little theorem. $\endgroup$ Oct 6, 2018 at 9:22
  • $\begingroup$ Yes, the process is fine. Just try to apply Fermat's theorem and Euler's theorem to find the remaining ones. $\endgroup$ Oct 6, 2018 at 9:24
  • $\begingroup$ With Fermat's Little Theorem I get that $17^{17} \equiv 17^5 \pmod{7}$, but I do not know how to proceed from there. $\endgroup$ Oct 6, 2018 at 9:25
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    $\begingroup$ You still have $17 \equiv 3 \pmod{7}$ and with Fermat's little theorem $17^{17}\equiv 17^5 \equiv 3^5 \equiv 3^2 \cdot 3^2 \cdot 3 \equiv 2\cdot2\cdot3 \equiv 5 \pmod{7}$ $\endgroup$
    – rtybase
    Oct 6, 2018 at 9:31
  • $\begingroup$ Multiply by $\dfrac 33$ so you have $\dfrac 13$ and the inverse of $3$ is clearly $5$. $\endgroup$
    – Piquito
    Oct 6, 2018 at 10:23

2 Answers 2

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Euler's Theorem states that $$ a^{\phi(n)}\equiv1 \pmod{n}$$ For all $n$, where $\phi(n)$ is the Euler totient function, denoting the number of positive integers less than $n$ relatively prime to $n$. For primes, $\phi(n)=n-1$. Thus $$ a^{n-1}\equiv1\pmod n. $$ This result is also known as Fermat's little theorem. Now, $7$ is prime, so anything to the power of $6$ is congruent to $1$. We have $$\begin{split} &15^{15}+16^{16}+17^{17}+18^{18}+19^{19}+20^{20}\\ \equiv\, & 15^3 + 16^4 + 17^5 + 18^0 + 19^1 + 20^2 \\ \equiv \, & 1^3 + 2^4 + 3^5 + 4^0 + 5^1 + 6^2\\ \equiv\, & 1+16+243+1+5+36\\ \equiv\, & 302\\ \equiv\, & 1\pmod{7}. \end{split}$$

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    $\begingroup$ You have made two minor mistakes. Firstly, it is false that Fermat's little theorem gives us that anything to the power of 6 is congruent to 1. The base that is raised to the power of 6 also cannot be divisible by 7 for it to always be congruent to 1. Secondly, it should say $2^4$ rather than $2^3$ on line three in your calculation. This yields the different, and correct, answer of the the sum in question being congruent to 1 modulo 7. $\endgroup$ Oct 6, 2018 at 10:34
  • $\begingroup$ For the first issue, it's only not true if $p\mid a$, which obviously isn't a problem here. You're right about the second part, it's a careless typo; I have fixed the issue. $\endgroup$
    – YiFan Tey
    Oct 6, 2018 at 10:39
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You can go with the same process I think:

$$17^{17} \equiv 3^{17} \mod{7}$$

and we have

$$3^1 \equiv 3 \mod{7}$$ $$3^2 \equiv 2 \mod{7}$$ $$3^3 \equiv 6 \mod{7}$$ $$3^4 \equiv 4 \mod{7}$$ $$3^5 \equiv 5 \mod{7}$$ $$3^6 \equiv 1 \mod{7} \text{ (We could have this by using Fermat's Little Theorem as well)}$$

From here, we can conclude that $3^{17} \equiv 5 \mod{7}$.

Then, notice that

$$18 \equiv 4 \equiv -3 \mod{7}$$ $$19 \equiv 5 \equiv -2 \mod{7}$$ $$20 \equiv 6 \equiv -1 \mod{7}$$ which means you can find other terms by using your previous results for $15,16$ and $17$.

But as suggested on comments, a better way is to use Fermat's Little Theorem.

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    $\begingroup$ If you note the sum of bases is $15+16+17+18+19+20=105=15\times 7$. Also the sum of remainders of bases divided by 7 is $1+2+3+4+5+6=21=3\times 7$. In this case the sum is divisible by 7, that is the remainder is 0. This is a particular specification of number 7. if you add another term like $21^{21}$ or $28^{28}$, the result will be the same. $\endgroup$
    – sirous
    Oct 6, 2018 at 11:00

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