help I have question about squares

Let $a$ and $b$ be non-zero integers. If both $a^2$ and $b^2$ have at most three digits, then what is the greatest possible difference between $a^2$ and $b^2?$

maximum Aug 14, 2023

#1**+1 **

Let's analyze the problem step by step:

We are given that both $a^2$ and $b^2$ have at most three digits. This means that the largest possible value for a three-digit number is $999$, and the smallest possible value for a three-digit number is $100$. So, we have the range for $a^2$ and $b^2$ as $100 \leq a^2, b^2 \leq 999$.

To maximize the difference between $a^2$ and $b^2$, we should maximize $a^2$ and minimize $b^2$. The largest possible value for $a^2$ is $999$ (since it's at the upper limit of the range), and the smallest possible value for $b^2$ is $100$ (since it's at the lower limit of the range).

So, the greatest possible difference between $a^2$ and $b^2$ is:

$$\text{Difference} = a^2 - b^2 = 999 - 100 = 899.$$

Therefore, the greatest possible difference between $a^2$ and $b^2$ is $899$.

SpectraSynth Aug 14, 2023

#2**+1 **

Note that both a^2 and b^2 have* AT MOST* three digits

But this means that a^2 could have three digits but b^2* could *have only one digit

Since both a and b are non-zero integers the largest that a could be = 31 so that 31^2 = 961

And the smallest that b could be = 1 so that 1^2 = 1

So the greatest difference =

a^2 - b^2 = 961 - 1 = 960

CPhill Aug 14, 2023