Can this equation be solved using integer numbers?

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The equation 3^x = 4y + 5 can be analyzed by expressing one variable in terms of the other to explore integer solutions. It is suggested to consider the equation modulo 4, leading to the condition 3^x ≡ 1 (mod 4). This relationship is connected to Fermat's Little Theorem, which indicates that x must be a multiple of 3 for integer solutions to exist. The discussion revolves around how to manipulate the equation to isolate variables effectively. Ultimately, the focus is on determining the conditions under which integer solutions are possible.
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How to solve that eqation?
3^x=4y+5
 
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Express one variable in terms of the other (see which makes most sense). What are the conditions for having integer solutions?
 
That is the same as saying that 3^x= 1 (mod 4). Check out "Fermat's Little Theorem".
 
Ok, but how can I show that this equation can be saying like that you wrote? If I place for 3^x=1(mod 4) to top equation and from Fermat's little theorem I have x=3 than I have equation with one variable? That you thought about?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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