Priya's age was equal to the cube of an integer (other than 1) four years ago and will be equal to the square of an integer four years from now. After how many more years will her age be such that one year earlier it was a perfect square and one year later it will be a perfect cube?

Introduction / Context:
This is a higher level problem on ages combined with number theory. Priya's age is related to perfect cubes and perfect squares at different times. Four years ago her age was a perfect cube, four years from now it will be a perfect square, and we must find when her age will lie exactly between a perfect square and a perfect cube in consecutive years. This type of puzzle tests both algebraic reasoning and familiarity with perfect powers.

Given Data / Assumptions:
Four years ago, Priya's age was equal to the cube of an integer greater than 1. Four years from now, her age will be equal to the square of an integer. We must find how many more years she should wait so that at that future time her age will have the property that the age one year earlier is a perfect square and the age one year later is a perfect cube. Ages are in whole years.

Concept / Approach:
We represent Priya's present age by a variable and convert the conditions into equations involving perfect powers. First, we search for an age x such that x - 4 is a perfect cube and x + 4 is a perfect square. Once we find such an x, we then look for a future age A = x + k such that A - 1 is a perfect square and A + 1 is a perfect cube. This can be approached either by systematic trial with small perfect cubes and squares or by logical reasoning with typical age ranges.

Step-by-Step Solution:
Step 1: Let Priya's present age be x years.Step 2: Four years ago, her age was x - 4 and this is given to be a perfect cube greater than 1. Try small cubes: 2^3 = 8, 3^3 = 27, and so on. So x - 4 could be 8, 27, etc.Step 3: Four years from now, her age will be x + 4 and this must be a perfect square, such as 9, 16, 25, 36, and so on.Step 4: Testing x - 4 = 8 gives x = 12. Then x + 4 = 16, which is 4^2, a perfect square. So x = 12 satisfies both conditions: 4 years ago her age was 8 (2^3) and 4 years from now it will be 16 (4^2).Step 5: Now we must find k such that at age A = x + k, the age one year earlier, A - 1, is a perfect square and the age one year later, A + 1, is a perfect cube.Step 6: With x = 12, test reasonable future ages. Try A = 26: then A - 1 = 25 = 5^2 (square) and A + 1 = 27 = 3^3 (cube). So A = 26 years works.Step 7: The waiting time is k = A - x = 26 - 12 = 14 years.

Verification / Alternative check:
Check the initial conditions: Priya is now 12 years old. Four years ago, she was 8 years old, which is the cube of 2. Four years from now, she will be 16 years old, which is the square of 4. After waiting 14 more years, she will be 26 years old. At that time, the previous year's age, 25, is 5^2 and the next year's age, 27, is 3^3. All conditions are perfectly satisfied.

Why Other Options Are Wrong:
If she waited 7 or 12 years, the resulting ages would not simultaneously satisfy the condition that one year earlier is a perfect square and one year later is a perfect cube. For example, at 19 or 24 years, the neighbouring ages do not form the required perfect powers. Similarly, waiting 21 years would lead to an age where neighbouring ages are not one perfect square and one perfect cube. Only 14 years gives the exact combination needed.

Common Pitfalls:
Many learners try random guesses for her present age without using the combined constraints of a cube and a square at four-year intervals. Another common mistake is misreading the final condition and thinking that her age itself must be both a square and a cube, which is not possible except for special numbers. Here, her age lies between a square and a cube in consecutive years, which is a subtler requirement.

Final Answer:
Priya should wait 14 years for the described condition to occur.