Discarding duplicate

We can actually solve this problem with pen and paper.

I won't explain the solution since jorgbrown has already done a great job in his post:

Suppose \( a \) is a perfect square of the smaller \( a \), but not a square of a square. Then we have a duplicate when \( b \) is \( 2, 3, 4\dots \) up to \( 50 \). That is, \( 49 \) duplicates.

Suppose \( a \) is a perfect cube of a smaller \( a \). When \( b \) is \( 2 \) through \( 33 \), we have duplicates of smaller \( a \) raised to the power \( b\times3 \). When \( b \) is \( 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66 \), we have duplicates of a smaller \( a \) raised to the power \( (\frac{b}{2})\times3 \). Total is \( 32 + 17 \), or again, \( 49 \) duplicates.

Suppose \( a \) is the square of the square of a smaller \( a \). When \( b \) is \( 2 \) through \( 49 \), we have duplicates of the square root of a raised to the power \( (b\times2) \).
When \( b \) is \( 51, 54, 57, 60, 63, 66, 69, 72, \) or \( 75, \) we have dupes of \( a^{(\frac{3}{4})} \) raised to the power \( \frac {b\times4} {3} \). Total is \( 49 + 9 \), or \( 58 \).

Suppose \( a \) is the fifth power of a smaller \( a \). We have dupes of fifth root of a raised to the power \( (b\times5) \), which covers \( b \) from \( 2 \) to \( 20 \). Then we have dupes of \( a^{(\frac{2}{5})} \) raised to the power \( \frac{b\times5}{2} \), which covers \( b \) of \( 22, 24, 26, 28, 30, 32, 34, 36, 38, 40 \). Then we have dupes of \( a^{(\frac{3}{5})} \) raised to the power \( \frac{b\times5}{3} \), which covers \( b \) of \( 21, 27, 33, 39, 42, 45, 48, 51, 54, 57, 60 \). Last, we have dupes of \( a^\frac{4}{5} \) raised to the power \( \frac{b\times5}{4} \), which covers \( b \) of \( 44, 52, 56, 64, 68, 72, 76\), and \( 80 \). Total dupes: \( 48 \).

And the last power we have to worry about is \( 6 \). We have dupes of the square root of a raised to power \( (b\times2) \), which covers \( b \) from \( 2 \) to \( 50 \). Then we have dupes of the sixth root to the power \( \frac{b\times6}{4} \), which covers \( b \) of \( 52 \), \( 54, 56, 58, 60, 62, 64, 66 \). And last we have dupes of the sixth root to the power \( \frac{b\times6}{5} \), which covers \( b \) of \( 55, 65, 70, 75 \), and \( 80 \). Total dupes: \( 62 \).

Now let's put it all together:

squares: \( 4, 9, 25, 36, 49, 100 \): These \( 6 \) squares have \( 49 \) dupes each, \( 6 \times 49 \) = \( 294 \)

cubes: \( 8, 27 \): These \( 3 \) cubes have \( 49 \) duplicates each: \( 2 \times 49 = 98 \)

4th power: \( 16, 81 \). These \( 2 \) have \( 58 \) dupes each: \( 2 \times 58 = 116 \)

5th power: \( 32 \). This has \( 48 \) dupes.

6th power: \( 64 \): this has \( 62 \) dupes.

Total # dupes: \( 618 \). \( 9801-618 \) is \( 9183 \).