Answer:
The quantity of natural numbers between [tex]a^{2}[/tex] and [tex]c^{2}[/tex] is [tex]2\cdot (a + b) + 1[/tex].
Step-by-step explanation:
If [tex]a^{2}[/tex], [tex]b^{2}[/tex] and [tex]c^{2}[/tex] are consecutive perfect squares, then both [tex]a[/tex], [tex]b[/tex] and [tex]c[/tex] are natural numbers and we have the following quantities of natural numbers:
Between [tex]b^{2}[/tex] and [tex]c^{2}[/tex]:
[tex]c^{2} = (b+1)^{2}[/tex]
[tex]c^{2} = b^{2}+2\cdot b + 1[/tex]
[tex]c^{2}-b^{2} = 2\cdot b + 1[/tex]
And the quantity of natural numbers between [tex]b^{2}[/tex] and [tex]c^{2}[/tex] is:
[tex]c^{2}-b^{2}-1 = 2\cdot b[/tex]
Between [tex]a^{2}[/tex] and [tex]b^{2}[/tex]:
[tex]b^{2} = (a + 1)^{2}[/tex]
[tex]b^{2} = a^{2} +2\cdot a + 1[/tex]
[tex]b^{2}-a^{2} = 2\cdot a + 1[/tex]
And the quantity of natural numbers between [tex]a^{2}[/tex] and [tex]b^{2}[/tex] is:
[tex]b^{2}-a^{2}-1 = 2\cdot a[/tex]
And the quantity of natural numbers between [tex]a^{2}[/tex] and [tex]c^{2}[/tex] is:
[tex]Diff = 2\cdot a + 2\cdot b + 1[/tex]
Please observe that the component +1 represents the natural number [tex]b^{2}[/tex]
