Consider two positive integers x and y, and determine all possible pairs (x,y) that satisfy this equation:
x
3 = 3
y - 1
Provide valid reasoning for your answer.
Rearrange the equation a bit:
x^3 - 1 = 3^y
(x-1) * (x^2+x+1) = 3^y
(x-1) * [(x-1)^2 + 3*(x-1) + 3)] = 3^y
Both factors must be powers of 3, assume that x-1 is a nontrivial power of 3, say 3^z with z>=1. Substitute this and then we get:
3^z * [3^(2z)+3^(z+1)+3] = 3^y
3^(z+1) * [3^(2z-1)+3^z+1] = 3^y
But now we have a problem. The larger factor in the last equation is now congruent to 1 mod 3. This contradicts the need for both factors to be powers of 3. So x-1 cannot be a nontrivial power of 3.
Then x-1 must equal 1 (=3^0), the trivial power of 3. Then x=2 and y=2 does form a solution and is the only positive integer solution to the equation.
Solution
