Question

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

Answer 1

Answer:

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

Step-by-step 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 $b^{2}$