How do you solve $\left( \frac{3^{2020}+5^{2020}+7^{2020}}{3} \right)^{2020}\mod 8 $

Question

is the answer simple, i look it up on wolfram alpha and i get complex result..

Answer 1

This is a more pedestrian answer, that ignores the knowledge of the ring $\Bbb Z/8$ and the fact that division with $3$ is allowed in it, and of the ring/field $\Bbb Z/3=\Bbb F_3$.

We have $$ 3^{2020} + 5^{2020} + 7^{2020} \\ = 9^{1010} + 25^{1010} + 49^{1010} \\ = (8+1)^{1010} + (3\cdot 8+1)^{1010} + (6\cdot 8+1)^{1010} \ . $$ Now we imaginary open the parentheses, getting only terms. All but one of the terms for each parenthesis are not divisible by $8$, these are $1^{1010}$, and $1^{1010}$, and $1^{1010}$. So the result is giving the rest $3$ modulo eight.

On the other hand, the first parenthesis is a power of $3$, thus divisible by $3$, and in $(3\cdot 8+1)^{1010}$, $(6\cdot 8+1)^{1010} $ only the terms $1^{1010}$, and $1^{1010}$ are not divisible by $3$.

So the number $$ 3^{2020} + 5^{2020} + 7^{2020} $$ has the rest $2$ when divided by three. The big expression $$ \left( \frac{3^{2020}+5^{2020}+7^{2020}}{3} \right) $$ ist thus a rational number with denominator $3$, which is irreducible. Taking the whole at power $2020$ introduces the denominator $3^{2020}$, and it cannot be taken modulo eight, this operation does not make any sense (without giving one to it).

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Note: If we know how to work in $\Bbb F_3$, then in this field $$ 3^{2020} + 5^{2020} + 7^{2020} = 0^{2020} + (-1)^{2020} + 1^{2020} = 0+1+1\ . $$

Answer 2

Note that $3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \pmod{8}$, so your expression simplifies to $\left ( \frac{1^{1010}+1^{1010}+1^{1010}}{3} \right )^{2020} \equiv 1^{2020} \equiv 1 \pmod{8}$.

Answer 3

Are you sure $5^{2020} + 7^{2020}$ is divisible by $3$? In my knowledge, it's not.