If $n\in\mathbb N$ and $k\in\mathbb Z$, solve $n^3-32n^2+n=k^2$.

Question

If $n\in\mathbb N$ and $k\in\mathbb Z$, solve $$n^3-32n^2+n=k^2$$

I've tried checking congruence modulo $4$. I have that n is divisible by 4. Let $n=4u$, where $u\in\mathbb N$. Then the equation becomes $$16u^3-32u^2+u=(\frac{k}{2})^2$$ And we know that $k$ is even (divisible by $4$). So $$16u^3-32u^2+u=r^2$$ Where $r\in\mathbb Z$. It is an even number as well. That's what I've tried so far.

Answer 1

First of all, let's factor the expression on the left-hand side:

$$n^3-32n^2+n = n(n^2-32n+1) = n\left((n-16)^2 - 255\right)$$

Since $\gcd(n,n^2-32n+1)=1$, both $n$ and $n^2-32n+1$ must be squares in order for their product to be a square. Thus, there is some integer $m$ such that $$(n-16)^2 - 255 = m^2$$ or, equivalently, $$(n-16-m)(n-16+m) = 255$$

Since the quantities on the left-hand side are integers, they must be divisors of $255$. If $d$ is any divisor of $255$, we'd have $$n = 16+\frac{1}{2}\left(d+\frac{255}{d}\right)$$

and we know that $n$ must be square itself. This leaves us with just eight possible values of $d$ to try ($\pm 1, \pm 3, \pm 5, \pm 15$; the remaining eight will be covered by them). As it turns out, only $d=(-15)$ and $d=1$ produce square value of $n$ ($n=144$ or $n=0$), which leads to three (or two, if we restrict ourselves to strictly positive values of $n$) solutions: $$(n,k)\in \{ (0, 0), (144, \pm 1524) \}$$.

Answer 2

If $0\in\mathbb N$, note that $n=k=0$ is a solution. For the rest of the argument assume $0\notin \mathbb N$.

If $n$ is odd, then $n^2\equiv 1\pmod 4$ and hence $n^3-32n^2 +n\equiv 2n\equiv 2\pmod4$, but $k^2$ can only be $\equiv 0\pmod 4$ or $\equiv 1\pmod 4$. Thus $n$ is even. Assume $2^r\|n$ (that is $2^r\mid n$, but $2^{r+1}\not\mid n$). Then we have just seen $r>0$ and thus we have $2^{r+1}\mid n^3-32n^2$ and hence $2^r\|k^2$. We conclude that $r$ is even. Let $p$ be an odd prime divisor of $n$, say $p^m\|n$ with $m>0$. Then again $p^m\|n^3-32n^2+n$ so that $m$ is even. We conclude that $n$ is a square. With $n=m^2$, the left hand side is divisible by $m^2$, so we $k$ is a multiple of $m$. Writing $k=ml$ the equation becomes (after division by $n=m^2\ne0$) $$ m^4-32m^2+1=l^2$$ If $m>12$, we have $$ m^4-34m^2+289<m^4-32m^2+1<m^4-32m^2+256,$$ that is $$ (m^2-17)^2<l^2<(m^2-16)^2,$$ which is absurd. The remaining cases $m=1,\ldots, 12$ can be checked manually and give us precisely two solution: $m=12$, $l=\pm127$. Substituting back, we obtain $$ n=144, k=\pm 1524. $$