This is a problem with multiple parts. When you complete a part and are told that your answers are correct then click to go on to the next part and resubmit your answer. When you do this ignore the statement that there are unanswered questions and just answer the new questions.

Here we are going to explore something that you will probably find very mysterious.

First, think about the areas of squares. There is certainly a square of area 1 square unit,
since we know that if we make its sides each 1 unit then its area must be $1\times 1$ square unit.
Similarly there is certainly a square of area 4 square units since its sides would have length 2 units.
Now suppose we gradually shrink the lengths of the sides of a square of area 4 square units
until we reach size 1 square unit.
Then, do you believe that somewhere during the shrinking we will have a square of area exactly 2 square units?

Most people believe that there is such a square. If there is, then its sides have length $\sqrt{2}$ units.
Now the ancient Greeks, who were the first Westerners to study geometry
first thought that every length was some fraction of a unit. Let us see whether that is true.

Suppose $\sqrt{2}$ is a fraction. Then, since every fraction can be reduced to lowest terms
we can say that

$\sqrt{2}=\frac{p}{q}$ where $\frac{p}{q}$ is reduced to lowest terms,
that is, $\ p\ $ and $\ q\ $ are integers and no integer greater than 1 divides both$p$ and $q$.

Now if two numbers are equal so are there squares so if $\sqrt{2}=\frac{p}{q}$
then $2= (\frac{p}{q})^2= \frac{p^2}{q^2}$.
But this means $p^2=2\cdot q^2$ and so $\ p^2\ $ must be even since 2 times any integer is an even integer.

Now, I will ask a little help from you. Fill in the following information.

$0^2$ =
$1^2$ =
$2^2$ =
$3^2$ =
$4^2$ =
$5^2$ =
$6^2$ =
$7^2$ =
$8^2$ =
$9^2$ =

Now, without squaring it out, you can see that the last digit of
$283859908828888997687654867987617^2$ must be