Question

A²,b²,c² are consecutive perfect squares.How many natura numbers are lying between a² and c², if a>0

Answer 1

The quantity of natural numbers between
`a^(2)` and
`c^(2)` is
`2\cdot (a + b) + 1`.

Explanation:

If
`a^(2)`,
`b^(2)` and
`c^(2)` are consecutive perfect squares, then both
`a`,
`b` and
`c` are natural numbers and we have the following quantities of natural numbers:

Between
`b^(2)` and
`c^(2)`:

`c^(2) = (b+1)^(2)`

`c^(2) = b^(2)+2\cdot b + 1`

`c^(2)-b^(2) = 2\cdot b + 1`

And the quantity of natural numbers between
`b^(2)` and
`c^(2)` is:

`c^(2)-b^(2)-1 = 2\cdot b`

Between
`a^(2)` and
`b^(2)`:

`b^(2) = (a + 1)^(2)`

`b^(2) = a^(2) +2\cdot a + 1`

`b^(2)-a^(2) = 2\cdot a + 1`

And the quantity of natural numbers between
`a^(2)` and
`b^(2)` is:

`b^(2)-a^(2)-1 = 2\cdot a`

And the quantity of natural numbers between
`a^(2)` and
`c^(2)` is:

`Diff = 2\cdot a + 2\cdot b + 1`

Please observe that the component +1 represents the natural number
`b^(2)`