This is in response to your question about the Day 16: Your Turn Explanation. I'm glad you asked, because the explanation is very interesting and a little abstract.

When we discussed the last digit of a number, and a square, we were really talking about the remainder that we get after dividing that number by $$10.$$ Now we are extending that idea to the case where instead of dividing by $$10,$$ we are dividing by $$4.$$

Let's look back to my comment in the Day 16: Challenge Discussions.

In the same way that before we only cared about the last digit squared, and ignored the $$100 \times \text{ something}$$ and $$10 \times \text{ something }$$ terms, here we only care about the remainder after dividing by $$4,$$ so we can ignore the multiples of $$4,$$ multiples of $$4^2,$$ multiples of $$4^3,$$ etc.

We want to look at square numbers and the remainders that we get after dividing these squares by $$4.$$ The clever trick is to look at just the remainders after dividing by $$4$$:

$$(4 \times \text{ something } + R)^2 = 16 \times \text{ something } + 4 \times \text{ something } + R^2$$

Something to be careful here is that the $$R$$ isn't any single-digit number. It's only $$0, 1, 2 \text{ or } 3,$$ because once we get $$R = 4,$$ then it gets incorporated into the $$4 \times \text{ something }$$ term instead.

So since $$R$$ can only be $$0, 1, 2, \text{ or } 3,$$ that means $$R^2$$ can only be $$0^2, 1^2, 2^2$$ or $$3^2.$$ These numbers are $$0, 1, 4$$ or $$9.$$ Since we care only about the remainder after dividing by $$4,$$ let's remove $$4$$ from the last two numbers. This means that $$R^2$$ can be $$0, 1, 0,$$ or $$1.$$

Since we found that squares are equal to $$4 \times \text{ something } + R^2,$$ and the first term is divisible by $$4,$$ then the remainder of $$R^2$$ (after dividing by $$4$$) is the remainder of the square number (after dividing by $$4$$). This means that squares either have remainders of $$0$$ or $$1$$ after dividing by $$4.$$

The question asks us to add together two squares. The possibilities for the remainder of this sum (after dividing by 4) are:

$$0 + 0 \\\\ 0 + 1 \\\\ 1 + 1 \\\\$$

which give us $$0, 1$$ or $$2.$$ It's impossible to get $$3.$$ The question asks us for the impossible remainder, so the answer is $$\boxed{3.}$$

I hope this helped! I'm really happy to answer any questions you have about the course content, no matter how big or small. I hope that you found this course beneficial to you and wish to learn more with us!

Happy Learning,

The Daily Challenge Team