Prove $5a^2\equiv k \pmod{12}$, where $k\in \{0,5,8,9\}$. Hence show that the equation $24x^7 + 5y^2 = 15$ has no integer solutions.
My lecturer used a table containing $a$, $a^2$, and $5a^2$ from $1$ to $11$ to which he applied $\pmod{12}$ for ex.
$a=3,a^2=9$, and $5a^2=9$
I didn't really understand what he continued on to do, does anyone have alternative method/proof?
On
Not sure if it's a typo or you copied the problem incorrectly. You mention mod $12$ below and it looks like it should be the same above. A square should be equivalent to $0$ or $1$ mod $3$ and $4$, which (when multiplied by 5) is what that set seems to represent. Also, when you take your equation and consider the equivalence mod $12$, the first term now drops out leaving
$$5y^2\equiv3\pmod{12}$$
which is not on your list.
On
The squares modulo $\;12\;$ are
$$0^2=\color{red}0\;,\;1^2=\color{red}1\;,\;2^2=\color{red}4\;,\;3^2=\color{red}9\;,\;4^2=\color{red}{4}\;,\;5^2=\color{red}1\;,\;6^2=\color{red}0,\;$$
(why the above are them all?), and from here
$$5a^2=\begin{cases}0\\5\\8\\9\end{cases}\pmod{17}$$
Working now your equation modulo $\;12\;$ we get
$$5y^2=3$$
which, as seen above, has no solution.
Your lecturer calculated the remainder when $5a^2$ is divided by $12$, for $a=0$ to $11$. Actually we can stop at $6$, since for any $a$ we have $(12-a)^2\equiv a^2\pmod{12}$. (Indeed we could stop at $a=3$, since $(6-a)^2\equiv a^2\pmod{12}$.)
So calculating from $0$ to $6$, we find that modulo $12$, $5y^2$ can take on values $0,5,8,9, 8, 5,0$.
Now suppose that $24x^n+5y^2=15$. Then, modulo $12$, we have $5y^2\equiv 3$. That is impossible, since $3$ is not in our list of possible remainders when $5a^2$ is divided by $12$.
Remark: It is simpler to work modulo $4$. If $(x,y)$ is a solution of our equation, then $5y^2\equiv 3\pmod{4}$. That cannot happen if $y$ is even. And if $x\equiv \pm 1\pmod{4}$, then $5x^2\equiv 1\pmod{4}$, contradicting the fact that $5y^2\equiv 3\pmod{3}$.