|Jul31-07, 12:00 PM||#1|
Subtract The Exponents, Get 3
Determine all possible non negative integer pairs (x, y) satisfying this equation:
2^x – 5^y = 3
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.
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