<s>[INST] Proving 7n+5 is never a cubic number? This is from a question that starts with:
An arithmetic progression of integers an is one in which $a_n=a_0+nd$, where $a_0$ and $d$ are integers and n takes successive values $0, 1, 2, \cdots$ Prove that if one term in the progression is the cube of an integer there will be an infinite number of such terms.
I have done this part (with help) but I am now stuck on proving that $7n+5$ can never be a cubic number (i.e. $x^3$ where $x$ is an integer) and $n$ is a positive integer. My first plan was proof by induction, trying this with both $x$ (from $x^3=7n+5$) and then $n$. Nether of these worked as I got formula that seemed impossible to solve. I have also tried manipulating $x^3=7n_1+5$ and $y^3=7n_2+5$ since if $x$ is an integer then there must also be a $y$ which this statement is also true. Any hints on where to start would be great thanks!
 [/INST] This is quite a quick thing to prove with modular arithmetic, but I will avoid that and use first principles. Suppose $m^3 = 7n+5$. There are seven cases:


*

*$m=7k$

*$m=7k+1$

*$m=7k+2$

*$m=7k+3$

*$m=7k+4$

*$m=7k+5$

*$m=7k+6$


So basically, $m=7k+i$ for $i$ in $\{0,1,\ldots,6\}$.
Now
$$\begin{align}
(7k+i)^3 &= 7n+5\\
7^3k^3+3\cdot7^2k^2i+3\cdot7ki^2+i^3&=7n+5\\
7^3k^3+3\cdot7^2k^2i+3\cdot7ki^2&=7n+5-i^3\\
7^3k^3+3\cdot7^2k^2i+3\cdot7ki^2-7n&=5-i^3\\
7\left(7^2k^3+3\cdot7k^2i+3\cdot k i^2-n\right)&=5-i^3
\end{align}$$
And this means $7$ is a divisor of $5-i^3$. If we check what $5-i^3$ is for all seven cases, we see that this is impossible. So the original premise that $m^3 = 7n+5$ cannot be true.
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