Solve the system of linear equations $x \equiv 5 \pmod{11}$ and $x \equiv 8 \pmod{24}$. Use the method demonstrated in class.

In the following problem, try to find a method of determining an example, not just trial and error. Give an example of integers $m \neq n$ which are NOT coprime (so $\gcd(n,m) > 1$), and integers $a$ and $b$ so that the system of linear equations $x \equiv a \pmod m$, $x \equiv b \pmod n$ has:

no solution

more than one solution (a different $a$ and $b$ will be needed)

explain your method for finding your examples

This problem is about square roots.

How many square roots does -1 have in the real numbers?

How many square roots does -1 have in the complex numbers?

Using Sage, find all the square roots of $-1$ modulo $5$, modulo $7$, modulo $13$ and modulo $5 \cdot 13=65$. Hint: you can use a for loop and check for every residue $x$ whether $x^2$ comes out to $-1$, and print it out when it does. Screenshot your code to include in what you submit. The Sage Sandbox should be good enough for this (a few lines of code).

Do you think this is weird?

Can you find a modulus where $-1$ has more than 4 square roots?

If there’s time remaining, consider whether the following statement is true or false: $a$ is invertible modulo $nm$ if and only if $a$ is invertible modulo $n$ and invertible modulo $m$. Maybe it is true sometimes (for some types of pairs $n$ and $m$).