Note that
$$n^2+20n+11=(n+10)^2-89$$
so we have to find all the pairs of squares $a^2$, $b^2$ such that $a^2-b^2=89$. Of course, $a$ and $b$ are integers. We see that $|a|>|b|$.
Since $a^2-b^2=(a+b)(a-b)$ and $89$ is a prime number, then we have these possibilities:
- $a+b=89$ and $a-b=1$, that is, $a=45$, $b=44$.
- $a+b=1$ and $a-b=89$, that is, $a=45$, $b=-44$.
- $a+b=-89$ and $a-b=-1$, that is, $a=-45$, $b=-44$.
- $a+b=-1$ and $a-b=-89$, that is, $a=-45$, $b=44$.
So clearly, $a^2=2025$ and $b^2=1936$.
Then $n+10=\pm45$ which gives two values for $n$, namely $n=35$ and $n=-55$.
EDIT: How can be proved that there are no more solutions?
Indeed, this is already done. Our reasoning is like this:
- Proposition $A$: "$n$ is such that $n^2-20n+11$ is a perfect square".
- Proposition $B$: "$n$ is $35$ or $-55$".
We have shown, assuming that $n$ is integer, that $A$ implies $B$. So if $n$ is another integer number, $B$ is false. Hence, $A$ is false. That is, if $n$ is not $35$ or $-55$ then $n^2+20n+11$ is not a perfect square.
The equation $y^2=1372x^4-3$ has only one positive integral solution for $x$ and $y$ at which is found at $(1,37)$.
We can use the general technique in this answer https://mathoverflow.net/a/338108 to convert your quartic into Weierstrass form and then we can use MAGMA to find all integral points on the curve.
Step 1: Quartic to Cubic (Weierstrass form)
$y^2=1372x^4-3$ can be transformed into $Y^2=X^3-4116X$ using $X:=1372x^2$ and $Y:=1372xy$ via the steps below
Take
$$y^2=1372x^4-3$$
Multiply both sides by $1372^2x^2$
$$1372^2x^2y^2=1372^3x^6-3\times1372^2x^2$$
$$(1372xy)^2=(1372x^2)^3-(3\times1372)(1372x^2)$$
$$Y^2=X^3-4116X$$
Step 2: Search for Integral Points
Then using MAGMA (An online version is here for you to confirm my work for yourself: http://magma.maths.usyd.edu.au/calc/) we can run the following two lines of code to find all of the integral points on our curve:
E := EllipticCurve([0,0,0,4116,0]);
IntegralPoints(E);
And we get the result: $(0 : 0 : 1)$ which tells us that the only one solution exists (the one that we found manually $(1,37)$).
Alternatively: Easier Solution
We could also run the following to get this answer directly (I realized this command existed after doing the work above, but it confirms the same answer).
IntegralQuarticPoints([1372, 0, 0, 0, -3]);
which gives the only positive output as $[ 1, 37 ]$
Best Answer
It always helps to form squares from the biggest power and is a good strategy: $$n^4-4n^3+14n^2-20n+10=n^2(n^2-4n+4)+10n^2-20n+10=\\ n^2(n-2)^2+10(n-1)^2=(n(n-2))^2+10(n-1)^2=((n-1)^2-1)^2+10(n-1)^2=\\ (n-1)^4-2(n-1)^2+1+10(n-1)^2=(n-1)^4+8(n-1)^2+1=\\ ((n-1)^2+4)^2-15=x^2$$ I think this quite large hint will make it a bit easier to solve it.
Just as a note: it just happens we can make a nice square, that is not always the case!