What is the solution space for the equation 2^x - 5^y = 3 in modular arithmetic?

[K Sengupta]

Problem

Determine all possible non negative integer pairs (x, y) satisfying this equation:

2^x – 5^y = 3


My Attempt:

If x =0, then 5^y = -2, which is a contradiction.

If x =1, then 5^y =-1, which is a contradiction.

If x = 2, then 5^y = 1, so that y = 0

If x>=3, then we observe that:

3^y = 3(Mod 8), so that y must be odd

Let us substitute y = 2s+1, where s is a positive integer. …….(*)

Again, if y – y, then 2^x = 4, giving: x = 2

If y =1, 2^x = 8, giving x = 3

For y>=3, substituting y = 2s+ 1 in terms of (*), we obtain:

2^x - 5^(2s+1) = 3
Or, 2^x = 3 (Mod 5)
or, x = 4t+3, where s is a non negative integer.

So we have:

2^(4t+3) – 5^(2s+1) = 3
Or, 8*(16^t) - 5*(25^s) = 3

For t =1, we obtain s =1, so that: (x, y) = (7, 3)

Hence, so far we have obtained (x,y) = (7, 3); (3,1) and (2, 0) as valid solutions to the problem.

**** I am unable to proceed any further, and accordingly, I am looking for a methodology giving any further valid solution(s) or any procedure conclusively proving that no further solutions can exist for the given problem.

 

[fresh_42]

Solve it modulo $3$.
\begin{align*}
&\phantom{\Longrightarrow} 2x-5y =3\\
&\Longrightarrow x\equiv y \mod 3 \\
&\Longrightarrow x=y+3k \\
&\Longrightarrow y=2k-1 \\
&\Longrightarrow x= 5k-1\\
&\Longrightarrow (x,y) =\{\,(5k-1,2k-1)\,|\,k\in \mathbb{Z}\,\}
\end{align*}