1. ## Congreunce modulo 49

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

2. By using the squared method:
$\displaystyle 6^{33}\equiv 6^{32}6\equiv 36^{16}6\equiv (-13)^{16}6$
$\displaystyle \equiv 13^{16}6\equiv 169^{8}6\equiv 22^{8}6\equiv 484^{4}6$$\displaystyle \equiv (-6)^{4}6\equiv 36^{2}6\equiv 13^{2}6\equiv 22\cdot 6\equiv 132\equiv 34 I leave it to you to find \displaystyle 8^{33}. 3. Entering this into my computer gives \displaystyle 6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right] which might be why you are having trouble 4. \displaystyle 8^{33}\equiv 36 (mod 49) thus we get 21 not 0. There may be other better method to prove this congruence equation. 5. Originally Posted by badgerigar Entering this into my computer gives \displaystyle 6^{33} + 8^{33} \equiv 21\left[ {\bmod 49} \right]$$\displaystyle$

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

6. Hello The general method is :

$\displaystyle \begin{array}{l} 8^0 \equiv 1\left[ {\bmod 49} \right] \\ 8^1 \equiv 8\left[ {\bmod 49} \right] \\ 8^2 \equiv 15\left[ {\bmod 49} \right] \\ 8^3 \equiv 22\left[ {\bmod 49} \right] \\ 8^4 \equiv 29\left[ {\bmod 49} \right] \\ 8^5 \equiv 36\left[ {\bmod 49} \right] \\ 8^6 \equiv 3\left[ {\bmod 49} \right] \\ 8^7 \equiv 1\left[ {\bmod 49} \right] \\ \end{array}$
The period is 7 then : n = 7k +p
but 33= 7×4+5
$\displaystyle 8^{33} \equiv 36\left[ {\bmod 49} \right]$
$\displaystyle same method : \begin{array}{l} 6^0 \equiv 1\left[ {\bmod 49} \right] \\ 6^1 \equiv 6\left[ {\bmod 49} \right] \\ 6^2 \equiv 36\left[ {\bmod 49} \right] \\ 6^3 \equiv 20\left[ {\bmod 49} \right] \\ 6^4 \equiv 22\left[ {\bmod 49} \right] \\ 6^5 \equiv 34\left[ {\bmod 49} \right] \\ 6^6 \equiv 8\left[ {\bmod 49} \right] \\ 6^7 \equiv 1\left[ {\bmod 49} \right] \\ \end{array}$
The period is 7 then : n = 7k +d

but 33= 7×4+5

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

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