Show that $(n^5-n+3)$ is not a perfect square

Question

I need to show that $n^5-n+3$ is not a perfect square. I think I have to use the Legendre symbol and quadratic residues, but I did not see how.

Instead I tried the following:

For $n=0,1$, we see that this is not a perfect square.

I observed that the distance between a square $k^2$ and $(k+1)^2$ is equal to $2k+1$, and the distance between $k^2$ and $(k-1)^2$ is equal to $2k-1$ for all $k>1$.

I then saw that $n^5=(n^3)^2$ is a perfect square for all $n$. Let now $k=n^3$. Then the closest square smaller than it is at least $2k-1=2n^3-1$ less than it.

We see that $n^5 - n + 3 = n^5 - (n - 3) = n^5 - (\sqrt[3]{k} - 3).$ We now need to show that $\sqrt[3]{k} - 3 < 2k - 1$. We get $\sqrt[3]{k} < 2k + 2$, which is true for all $k\geq 1$.

Therefore $n^5-n+3$ cannot be a perfect square.

Is this proof correct and am I right in assuming I could have used quadratic residues to do it more easily?

Answer 1

Your proof is wrong. $n^5\neq n^6=(n^3)^2$.

In the expression $n^5-n+3$, we see a power of $5$ in the exponent. That is excellent motivation to consider mod $5$ and look at quadratic residues there, since conveniently, $n^5\equiv n$ mod $5$ so the first two terms cancel out. This is Fermat's Little Theorem. Then, we are left with $n^5-n+3\equiv 3$ mod $5$, but $3$ is not a quadratic residue mod $5$ (I'm sure you know why).

Answer 2

Using Fermat's little $$n^5-n\equiv0\pmod5$$

$\implies n^5-n+3\equiv3$

Now for any integer $m,$

$$m\equiv0\pm1,\pm2\pmod5\implies m^2\equiv0,1,4$$ none of which $\equiv3\pmod5$