If $1835^2 \equiv 1 \mod{7}$ and $1986^3 \equiv 0 \mod{7}$, why is $1835^2 + 1986^3 \equiv 0 \mod{7}$

Question

I'm not sure what I'm missing; I've tried to apply Fermat's Little Theorem to prove that $1835^{1910} + 1986^{2061} \equiv 0 \mod{7}$.

Edit: OK, so my question now becomes: why is my calculator is telling me that $7 \mid 1986^3$?

input: 1986^3
output: 7833173256
input: ANS ÷ 7
output: 1119024751

Answer 1

You are probably using a calculator which rounds $\frac {1986^3}{7}$ to $1119024751$ and gives the false impression that $1986^3$ is a multiple of $7$

Do the calculation by hand and you see $\frac {1986^3}{7}$ is not an integer.

Answer 2

$\bmod 7\!:\,\ 1835\equiv 1,\ 1986\equiv\color{#c00}{ -2}\,$ (not $\,0),\,$ so

$\begin{align}\qquad\qquad &1835^{\large k}\! + 1986^{\,\large 3\,+\,6n}\\ \equiv &\ \ \ \ \ \ 1^{\large k}\! + (\color{#c00}{-2})^{\large 3(1+2n)}\\ \equiv &\ \ \ \ \ \ 1\ + (\,-1\,)^{\large\ 1+2n}\equiv 0\end{align}$

Edit $ $ Regarding the updated question, you don't need to use a calculator since it is easy mentally: $\!\bmod 7:\,\ 1986 = 2(10)^{\large 3}-14 \equiv 2(3^{\large 3})-0\equiv \color{#c00}{2(-1)}.\,$ In fact $\,10^{\large 3}\!+1 = 7\cdot 11\cdot 13\,$ so we can use $10^{\large 3}\equiv -1\,$ when computing mod any factor of $\,10^{\large 3}\!+1 = 1001,\ $ e.g. $\,77 = 7\cdot 11$.

Answer 3

$1986^3$ can't be $0$ because $1986\equiv 5\bmod 7$ and $\mathbf Z/7\mathbf Z$ is a field.

Namely, $1986^3\equiv 5^3\equiv -2^3\equiv -1\mod 7$ and $1835\equiv1\bmod 7$, so what is true is that $$1835^{1910}+1986^3\equiv 1+(-1)\equiv 0\mod 7.$$

Answer 4

$$1986 \equiv 5 ( mod \ 7) \\ 1986^{3} \equiv 5^{3} \equiv \color{#c00}{6} (mod \ 7) $$