How to solve that eqation?

$\displaystyle 3^x=4y+5$

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- Dec 19th 2010, 12:18 PMoszust001Solve in integer numbers
How to solve that eqation?

$\displaystyle 3^x=4y+5$ - Dec 19th 2010, 12:19 PMdwsmith
$\displaystyle 3^x\equiv 5=1 \ \mbox{(mod 4)}$

- Dec 19th 2010, 12:25 PMoszust001
Ok, but what it is. How can I do i to my equation.

- Dec 19th 2010, 12:26 PMdwsmith
What happens when x = 2? How about x = 4? ....

- Dec 19th 2010, 12:28 PMoszust001
Can you solve me a complete this equation, Step by step?

- Dec 19th 2010, 12:37 PMdwsmith
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. - Dec 19th 2010, 12:46 PMemakarov
$\displaystyle 3^{2k}=9^k\equiv 1^k=1\pmod{4}$, but $\displaystyle 3^{2k+1}=9^k\cdot 3\equiv 3\pmod{4}$.