<s>[INST] Knowing that $u_1 = 1, u_2 = 3$ and $u_{n + 2} = 2u_{n + 1} - u_n + 1, \forall n \in \mathbb Z^+$, prove that $4u_{n + 2}u_n + 1$ is a square number. 
Knowing that $$\large \left\{ \begin{align*} u_1 = 1&, u_2 = 3\\ u_{n + 2} = 2u_{n + 1} - u_n + 1&, \forall n \in \mathbb Z^+ \end{align*} \right.$$, prove that $\large 4u_{n + 2}u_n + 1$ is a square number.

Here are my thoughts.
$f\colon \mathbb Z \to \mathbb Z^+, u_n \mapsto v_n = 2u_n + 1, n \in \mathbb Z^+$. The problem becomes

Knowing that $$\large \left\{ \begin{align*} v_1 = 3&, v_2 = 7\\ v_{n + 2} + v_n = 2(v_{n + 1} + 1)&, \forall n \in \mathbb Z^+ \end{align*} \right.$$, prove that $\large v_{n + 2}v_n - v_{n + 1}$ is a square number.

This is a problem using induction, but I've just started studying the basics of it and haven't had any experience on the implications of it.
 [/INST] Use characteristic polynomials, like here, to get to the conclusion that
$$(x-1)(x^2-2x+1)=0$$
is the characteristic polynomial and $x=1$ is a root of multiplicity 3. Then
$$u_n=(an^2+bn+c)\cdot 1^n$$
and 
$$u_1=1=a+b+c$$
$$u_2=3=4a+2b+c$$
$$u_3=6=9a+3b+c$$
which resolves into $a=\frac{1}{2}$, $b=\frac{1}{2}$ and $c=0$, thus
$$u_n=\frac{1}{2}(n+1)n$$
and 
$$4u_{n+2}u_n +1=(n+3)(n+2)(n+1)n+1=\\
(n^2 + 3 n)^2 + 2 (n^2 + 3 n) + 1=\\
(n^2 + 3 n+1)^2$$
</s>