Answer:
9, 17, 3, 1600, 11
Step-by-step explanation:
Difference of two perfect squares:
[tex]a^2-b^2=(a-b)(a+b)[/tex]
It seems that you want some
[tex]x=a^2-b^2, x \in \mathbb{Z}_{\ge 0}[/tex]
Note: there's no official symbol for the set of whole numbers, I've already seem [tex]\mathbb{W}, \mathbb{Z}^+[/tex], as well.
[tex]5^2-4^2=25-16=9[/tex]
There are infinitely many numbers that can satisfy the condition given.
The only condition is that [tex]a\geq b[/tex], once we are considering whole numbers and not integers.
[tex]x_{1}=5^2-4^2=25-16=9[/tex]
[tex]x_{2}=9^2-8^2=81-64=17[/tex]
[tex]x_{3}=2^2-1^2=4-1=3[/tex]
[tex]x_{4}=50^2-30^2=2500-900=1600[/tex]
[tex]x_{5}=6^2-5^2=36-25=11[/tex]
