How to solve that eqation? $\displaystyle 3^x=4y+5$
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$\displaystyle 3^x\equiv 5=1 \ \mbox{(mod 4)}$
Ok, but what it is. How can I do i to my equation.
What happens when x = 2? How about x = 4? ....
Can you solve me a complete this equation, Step by step?
I pretty much solved it for you already. When x = 2, the congruence is true. When x = 4, it is true. x = 6, true as well. x = 8 guess what? True. Notice anything? Let's not forgot x = 0 is true too.
$\displaystyle 3^{2k}=9^k\equiv 1^k=1\pmod{4}$, but $\displaystyle 3^{2k+1}=9^k\cdot 3\equiv 3\pmod{4}$.
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