What is the first natural number that is a perfect square and a perfect cube? Can you find the second? The third? Explain.
so we want
a^2 = b^3
a = b^(3/2) or (√b)^3
so pick b = a perfect square,
e.g. b = 4 , then a = 8
so 8^2 = 4^3 = 64
let b = 9 , a = 27
9^3 = 27^2 = 729
let b = 16 , then a = 64
16^3 = 64^2 = 4096
let b = 25 , a = 125
25^3 = 125^2 = 15625
So the first few numbers with that property are:
64 , 729 . 4096 , 15625
As you can see, we can form as many of these as we want, there would be an infinite number of them.
Ok thank you. I understand now.
To find the first natural number that is both a perfect square and a perfect cube, we need to find a number that can be expressed as both x^2 (perfect square) and y^3 (perfect cube), where x and y are natural numbers.
To find such a number, we can start by listing perfect squares and perfect cubes and looking for any common values:
Perfect squares: 1, 4, 9, 16, 25, 36, ...
Perfect cubes: 1, 8, 27, 64, 125, 216, ...
By comparing the two lists, we see that the first natural number that is both a perfect square and a perfect cube is 1. Since 1 is the smallest natural number, it is the only number that meets this condition.
If we want to find the second natural number, we can continue this process by listing perfect squares and perfect cubes in increasing order. However, we can quickly observe that for any positive integer greater than 1, the power of 2 in its prime factorization will be twice the power of 3 (since squares have an even exponent and cubes have an odd exponent). As a result, there are no other natural numbers beyond 1 that can simultaneously be perfect squares and perfect cubes.
Therefore, the first natural number that is both a perfect square and a perfect cube is 1, and there are no additional natural numbers that satisfy this condition.