# Solve in integer numbers

• December 19th 2010, 12:18 PM
oszust001
Solve in integer numbers
How to solve that eqation?
$3^x=4y+5$
• December 19th 2010, 12:19 PM
dwsmith
$3^x\equiv 5=1 \ \mbox{(mod 4)}$
• December 19th 2010, 12:25 PM
oszust001
Ok, but what it is. How can I do i to my equation.
• December 19th 2010, 12:26 PM
dwsmith
What happens when x = 2? How about x = 4? ....
• December 19th 2010, 12:28 PM
oszust001
Can you solve me a complete this equation, Step by step?
• December 19th 2010, 12:37 PM
dwsmith
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.
• December 19th 2010, 12:46 PM
emakarov
$3^{2k}=9^k\equiv 1^k=1\pmod{4}$, but $3^{2k+1}=9^k\cdot 3\equiv 3\pmod{4}$.