Diophantine equation - Video explanation by Michael Penn I watched this video by Michael Penn:
https://www.youtube.com/watch?v=no7VzKmcyIg&ab_channel=MichaelPenn
The problem is find all $x,y,z \in \mathbb{N}$ such that:
$$ x^3 + 2y^3 + 4z^3 = 9!$$
The problem starts at about: 16:48 , however I have a question about the end of the solution..:
At the 33:33 he made a chart with $n, n^3 , 2n^3$ and he proved there is not combination of $n$ such that $2a^3 + c^3 = 52$

But my question is - who said $a$ and $c$ depend on $n$? this goes against my intuition of solving a Diophantine equation.. let's say we have this equation: $2a + c = 4$
So I know $c$ must be even - but using Michael's way I can just do this: $2n + n = 4 \Rightarrow 3n = 4 \Rightarrow \text{no natural solutions}$ But there are! $a = 1 , c = 2$ for example.
So I don't understand this argument of saying they both depend on the same variable..
Hope I was clear enough, thank you!
 A: There is no dependence on $n$ anywhere. Your teacher is just making a table of possible cubes and double cubes. Consider trying to solve $2a+2b=5$ in the natural numbers. The possible values of $2a$ are $2,4,6$ (as well as larger numbers) and the possible values of $2b$ are $2,4,6$ (as well as larger numbers). No combination of those possibilities gives $5$. So there is no solution.
That's what your teacher has done. He has $n$ in the table to make the table understandable, but that is all $n$ is used for. It is not inserted into the equation anywhere.
A: A quick recap of the argument: Repeatedly reducing mod $2$ and reducing mod $9$ shows that
$$(x,y,z)=(24a,12b,12c),$$
for some natural numbers $a$, $b$ and $c$, and hence the problem reduces to solving
$$4a^3+b^3+2c^3=3\cdot5\cdot7=105,$$
in natural numbers $a$, $b$ and $c$. In particular this implies that
$$4a^3,b^3,2c^3\leq105,$$
and so $a$, $b$ and $c$ must be rather small. If we make a list of small cubes and their relevant multiples:
$$\begin{array}{c|c|c|c}
n&n^3&2n^3&4n^3\\
\hline
0&0&0&0\\
1&1&2&4\\
2&8&16&32\\
3&27&54&\color{red}{108}\\
4&64&\color{red}{128}&\\
5&\color{red}{125}\\\end{array}$$
We see that we need not look any further as the (multiples of) cubes become too large. This shows that $a\leq 2$, $b\leq4$, and $c\leq3$. Now it is simply a matter of checking these few values of $a$, $b$ and $c$ to see whether we get a solution. This is the way in which the (similar) table is used in the video.
In the video the number of combinations of values to check is reduced by noting that $b$ must be odd, so either $b=1$ or $b=3$, yielding the equations
$$2a^3+c^3=52\qquad\text{ and }\qquad 2a^3+c^3=39,$$
respectively. We can again make a list of small cubes and their relevant multiples:
$$\begin{array}{c|c|c|c}
n&n^3&2n^3\\
\hline
0&0&0\\
1&1&2\\
2&8&16\\
3&27&\color{red}{54}\\
4&\color{red}{64}&\\
\end{array}$$
As before this now shows that $c\leq3$ and $a\leq2$, and a quick check shows that no combination of these values yields a solution.
