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.

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 
$$4u_{n+2}u_n +1=(n+3)(n+2)(n+1)n+1=\\
(n^2 + 3 n)^2 + 2 (n^2 + 3 n) + 1=\\
(n^2 + 3 n+1)^2$$
