1. ## Congreunce modulo 49

Prove that :
$6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]$

2. By using the squared method:
$6^{33}\equiv 6^{32}6\equiv 36^{16}6\equiv (-13)^{16}6$
$\equiv 13^{16}6\equiv 169^{8}6\equiv 22^{8}6\equiv 484^{4}6$ $\equiv (-6)^{4}6\equiv 36^{2}6\equiv 13^{2}6\equiv 22\cdot 6\equiv 132\equiv 34$
I leave it to you to find $8^{33}$.

3. Entering this into my computer gives
$6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]
$

which might be why you are having trouble

4. $8^{33}\equiv 36$ (mod 49)
thus we get 21 not 0.
There may be other better method to prove this congruence equation.

Entering this into my computer gives
$6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]$ $
$

which might be why you are having trouble
Hello thank you you have 21

6. Hello The general method is :

$\begin{array}{l}
8^0 \equiv 1\left[ {\bmod 49} \right] \\
8^1 \equiv 8\left[ {\bmod 49} \right] \\
8^2 \equiv 15\left[ {\bmod 49} \right] \\
8^3 \equiv 22\left[ {\bmod 49} \right] \\
8^4 \equiv 29\left[ {\bmod 49} \right] \\
8^5 \equiv 36\left[ {\bmod 49} \right] \\
8^6 \equiv 3\left[ {\bmod 49} \right] \\
8^7 \equiv 1\left[ {\bmod 49} \right] \\
\end{array}
$

The period is 7 then : n = 7k +p
but 33= 7×4+5
$8^{33} \equiv 36\left[ {\bmod 49} \right]
$

$
same method :
\begin{array}{l}
6^0 \equiv 1\left[ {\bmod 49} \right] \\
6^1 \equiv 6\left[ {\bmod 49} \right] \\
6^2 \equiv 36\left[ {\bmod 49} \right] \\
6^3 \equiv 20\left[ {\bmod 49} \right] \\
6^4 \equiv 22\left[ {\bmod 49} \right] \\
6^5 \equiv 34\left[ {\bmod 49} \right] \\
6^6 \equiv 8\left[ {\bmod 49} \right] \\
6^7 \equiv 1\left[ {\bmod 49} \right] \\
\end{array}
$

The period is 7 then : n = 7k +d

but 33= 7×4+5

$6^{33} \equiv 34\left[ {\bmod 49} \right]$
Conclusion :
$6^{33} + 8^{33} \equiv \left( {34 + 36} \right)\left[ {\bmod 49} \right]$

$6^{33} + 8^{33} \equiv \left( {34 + 36} \right)\left[ {\bmod 49} \right]because\left( {34 + 36} \right) \equiv 21\left[ {\bmod 49} \right]$