A To The Power B Plus C

It might not be the best solution, but here is how I did it:

Given that $c$ can be anything from $1$ to $9$ I started with the premise that the number I was looking for needed to be either $a^b$ itself or in the range where the difference between consecutive squares is greater than 9 .

This first occurs between $a=5$ and $a=6$ , where the difference between the squares is $6^2 - 5^2 = 11$ . We can add $10$ to $5^2$ to get our first possible answer, $35$ . You can quickly see though that 35 can be obtained by either $3^3 + 8$ or $2^5 + 3$ so that rules that one out. 36 can be also obtained a similar way.

So now we move up to the next range, between $a=6$ and $a=7$ . Adding $10$ to $6^2$ gets us $46$ , which you can validate pretty quickly as out of range of both $3^3$ and $2^5$ and all other powers available.

If b=1, we get numbers............. 2...to..10
If a=2........b=1 we get ...............3...to..11
.....................=2......."......................5...to..13
.....................=3......."......................9...to..17
.....................=4......."....................17...to..25
Next is 32+1=33 but we got only 25. So going for other power
...a=5...........=2......."....................26...to..34
...a=2...........=5......."....................33...to..41
Next 2^6 stars with 65, 3^3 =27 and we get.....28...to...36

So next would be 6^2=36........................""..........37...to...45
There is no way to get 46.

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