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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.

rtybase
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Lê Thành Đạt
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3 Answers3

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If you let $v_n = u_{n+1} - u_n$, then you have $v_1 = 2$ and $$v_{n+1} = v_n + 1$$ This leads to $v_n = n + 1$, so that $u_1 = 1$ and one has the recurrence $$u_{n+1} = u_n + n + 1$$ So $u_n = \sum_{k = 1}^n k = {n(n+1) \over 2}$. Hence $$4u_{n+2}u_n + 1 = (n+3)(n+2)(n+1)n + 1 $$ $$= (n^2 + 3n + 1)^2$$

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

rtybase
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Since your equation is linear, we can calculate that $$u_n=\frac{1}{2}n(n+1)$$, so $$4u_{n+2}u_n=n(n+1)(n+2)(n+3)+1=…$$

Dr. Sonnhard Graubner
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