Question

prove that there exists a pair of consecutive integers suchthat one of these integers is a perfect square and the otheris a perfect cube

Answer 1

To prove that there exists a pair of consecutive integers such that one is a perfect square and the other is a perfect cube, we can use the property that the difference between consecutive perfect squares and consecutive perfect cubes is always greater than 1.

Step-by-step explanation:

To prove that there exists a pair of consecutive integers such that one is a perfect square and the other is a perfect cube, we can use the property that the difference between consecutive perfect squares and consecutive perfect cubes is always greater than 1.

1. We start with an arbitrary perfect square, let's say n^2.
2. We then take the next consecutive perfect square, (n+1)^2 = n^2 + 2n + 1.
3. Finally, we take the cube root of the expression (n+1)^2 - n^2 and find that it is greater than 1, which means there exists a perfect cube between the two consecutive perfect squares.

Therefore, there exists a pair of consecutive integers, one of which is a perfect square and the other is a perfect cube.