Can you at least find one value each for a, b, and t so that this is true?
Hello, GGPaltrow!
I don't have the complete solution yet,
. . but I can narrow down the choices.
Note that will always end in 5.
Now note the final digit when adding
. .
The 4-step pattern repeats indefinitely.
Note that a square canNOT end in 7 or 3.
We find that 2 must be raised to an even power.
Does that help?
Let .
Then,
where p and q are non-negative integers.
Assume that p and q are both positive integers.
Then,
Consider the equation .
and
So, (Contradiction)
So, either p or q is zero. (Both p and q cannot be zero, because then b would be zero, which is not allowed by the question.)
ie either or is 1. Clearly, .
So,
Then, t can end with 3, 5, 7 or 9.
If t ends with 5, ends with 4 but then ends with 9 and will not be satisfied. So, t cannot end with 5. Similarly, t cannot end with 7 or 9.
So, t must end with 3. For t = 3, we have the solution (t, m, b) = (3, 1, 1), which leads to (a, b, t) = (2, 1, 3). I don't know if there are any more solutions.
Continuing alexmahone's argument, he has shown that
Subtract the second equation from the first to see that
But it is shown here that the equation has no solutions with m>0, b>1 and k=2. So we must conclude that b=1, and therefore (a,b,t) = (2,1,3) is the only solution.
Hmm.. When talking about this solution with my teacher, he said there's another one without involving such "advanced" things like Catalan's conjecture. Sooner or later he'll tell what he had in my mind but I wonder what this could have been.. Is there any such solution?