Let $A$ be the set of all integers of the form $ a^2 + 4ab + b^2$ where $a, b$ are integers
if $x, y$ are in $A$, prove that $xy$ is in $A$ (I have tried opening everything, it gets nowhere)
Show that $121$ is in $A$
Show that $11$ is not in $A$
Show that the equation $x^2 + 4xy + y^2 = 1$ has infinitely many integer solutions
Thank y'all for your help!
A clever trick (or a standard trick if you're familiar with using number fields to solve problems like these) to observe that
$$ a^2 + 4ab + b^2 = (a - b \alpha)(a - b \beta)$$
where $\alpha$ and $\beta$ are the two roots of the equation $x^2 + 4x + 1 = 0$.
Or, solutions given as $\frac{a}{b},$ beginning with a fake one with denominator zero, $$ \frac{1}{0},\frac{0}{1},\frac{-1}{4},\frac{-4}{15},\frac{-15}{56},\frac{-56}{209},\frac{-209}{780},\frac{-780}{2911}, $$
As you can see, both the sequence of numerators and the sequence of denominators satisfy $$ x_{n+2} = 4 x_{n+1} - x_n. $$
The way I found this was a bit intricate, but a proof, by induction, that these keep working is easy. By switching to a negative sign for convenience, so $$ a^2 - 4 a b + b^2 = 1 $$ we get (all) solutions as consecutive values in the sequence $$ -1,0,1,4,15,56,209,780,2911,\ldots $$
The reason they probably feel this is suitable for a contest is that the $4$ does not really matter: The solutions to $$ a^2 - 5 ab + b^2 = 1 $$ are consecutive numbers in $$ -1,0,1,5,24,115,551,2640,12649, \ldots $$ satisfying $$ x_{n+2} = 5 \, x_{n+1} - x_n. $$ For some positive $k > 2,$ the solutions to $$ a^2 - k \, ab+b^2 = 1 $$ are consecutive numbers in $$ -1,0,1,k, \ldots $$ satisfying $$ x_{n+2} = k \, x_{n+1} - x_n. $$
EddiTtT: It turns out this is not that pretty when considering the continued fraction for $\sqrt {k^2-4},$ partly because of the difference between odd and even $k.$ In the Lagrange cycle method that I prefer, it works very well: $\langle 1,k,1 \rangle$ is not reduced, but, with $k > 2,$ the equivalent form $\langle 1,k-2,2-k \rangle$ is reduced, and the entire Lagrange cycle is the shortest possible, length 2: $$ 0: \langle 1,k-2,2-k \rangle; \; \; \; 1: \langle 2-k,k-2,1 \rangle; \; \; \; 2: \langle 1,k-2,2-k \rangle; $$
Refresher course, why not. Given a discriminant $\Delta > 0$ but not a square, and integers such that $b^2 - 4 a c = \Delta,$ the form $\langle a,b,c \rangle$ refers to $$ f(x,y) = a x^2 + b x y + c y^2, $$ and this is called reduced if two conditions hold, $$ ac < 0 \; \mbox{and} \; \; b > |a+c|. $$ Probably not obvious, but these imply $|a|, b, |c| < \sqrt \Delta.$
This is only the solution to part 4. I assume that you have seen solutions to the rest. If not, I strongly recommend reading Ewan's answer.
Solution 1:
Apply Vieta's root jumping technique with a base case of $(4, -1)$. The result is immediate. This also explains why in Will's answer, you see that the $b$ and $a$ have the same (absolute) value. Note that you can multiply $a$ and $b$ by -1 and still get a solution.
Solution 2: In Ewan's answer, we get that
$$ f(ac−bd,ad+4bd+bc)=f(a,b)f(c,d). $$
which shows you how to 'multiply' solutions. Now, observe that $(4, -1)$ is a solution.
Claim: $ (4, -1)^n$ yields infinitely many distinct solutions.
Proof: You can easily show that $ ( 4, -1)^n \sim (4^n, ??)$
If you are familiar with Pell's Equation, Ewan draws the connection with the equation of the form $ x^2 - 3y^2 = 1$, which further explains / motivates why we likely can get all solutions through multiplication. It is not surprising that we reach a linear recurrence like that given in Will's answer, since the original Pell's equation also satisfies a linear recurrence.
1.) We can see that, $a^2 + b^2 + 4ab = (a+2b)^2 - 3b^2 = u^2 - 3v^2$. Also, $u$ and $v$ can take all integer values. Now, $u^2 - 3y^2 = (u + \sqrt 3)(u - \sqrt 3)$. So, $A = \{\,N(\alpha)\,\mid\,\alpha\in\mathbb Z(\sqrt 3)$. We know that $N(\alpha\beta) = N(\alpha)N(\beta)$, so if $x = N(\alpha) \in A$ and $y = N(\beta) \in A$, then $z = N(\alpha) \in A$.
4.) Now if $N(\alpha) = 1$, then $N(\alpha^k) = 1\,\forall k \in \mathbb N$. Consider $\alpha = 2 + \sqrt 3$, $|\alpha| > 1$, hence if $m < n \implies |\alpha^m| < |\alpha^n|, \implies \alpha^m \neq \alpha^n$. So $B = \{\,\alpha^n\,\mid\,n\in\mathbb N\}$, is an infinite solution set of $N(\alpha) = 1$.
