Maths Olympiad Prep

Track / Stage 6 / 128 of 400 #1128 of 1964

Problem 1128

National olympiad, first round
Algebra Difficulty 6.2 Prove it

Example 8 Define the sequence {an}:a1,a2\left\{a_{n}\right\}: a_{1}, a_{2} are the two roots of the equation z2+iz−1=0z^{2}+\mathrm{i} z-1=0, and for n⩾2n \geqslant 2, we have (an+1an−1−an2)+i(an−1+an−1−2an)=0\left(a_{n+1} a_{n-1}-a_{n}^{2}\right)+\mathrm{i}\left(a_{n-1}+a_{n-1}-2 a_{n}\right)=0.
Prove that for all positive integers nn, an2+an+12+an+22=anan+1+an+1an+2+an+2ana_{n}^{2}+a_{n+1}^{2}+a_{n+2}^{2}=a_{n} a_{n+1}+a_{n+1} a_{n+2}+a_{n+2} a_{n}.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Analyzing from the recursive formula, we derive the general term formula of the sequence {an}\left\{a_{n}\right\}, and then prove it.
Proof: From the equation z2+iz−1=0z^{2}+\mathrm{i} z-1=0, we can solve for z=−i±32=−1±3i2⋅i=iwz=\frac{-\mathrm{i} \pm \sqrt{3}}{2}=\frac{-1 \pm \sqrt{3} \mathrm{i}}{2} \cdot \mathrm{i}=\mathrm{i} w or iw2\mathrm{i} w^{2},
where w=−12+32iw=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}.
Without loss of generality, let a1−iw2,a2−iwa_{1}-\mathrm{i} w^{2}, a_{2}-\mathrm{i} w. From the known recursive relation, we have
an2+2ian+i2=an+1an−1+i(an+1+an−1)+i2,  a_{n}^{2}+2 \mathrm{i} a_{n}+\mathrm{i}^{2}=a_{n+1} a_{n-1}+\mathrm{i}\left(a_{n+1}+a_{n-1}\right)+\mathrm{i}^{2} \text {, }

which simplifies to (an+i)2=(an+1+i)(an−1+i)\left(a_{n}+\mathrm{i}\right)^{2}=\left(a_{n+1}+\mathrm{i}\right)\left(a_{n-1}+\mathrm{i}\right).
Clearly, an+i≠0(n∈N∗)a_{n}+\mathrm{i} \neq 0\left(n \in \mathbf{N}^{*}\right). Otherwise, if there exists some n∈N∗n \in \mathbf{N}^{*} such that an+i=0a_{n}+\mathrm{i}=0, then from (1) we get an−1+i=0a_{n-1}+\mathrm{i}=0, and continuing this, we have a2+i=0a_{2}+\mathrm{i}=0, which contradicts a2=iwa_{2}=\mathrm{i} w. Therefore, the sequence {an+i}\left\{a_{n}+\mathrm{i}\right\} is a geometric sequence with the first term a1+i=i(w2+1)a_{1}+\mathrm{i}=\mathrm{i}\left(w^{2}+1\right) and the common ratio a2+ia1+i=iw+iiw2+i=w\frac{a_{2}+\mathrm{i}}{a_{1}+\mathrm{i}}=\frac{\mathrm{i} w+\mathrm{i}}{\mathrm{i} w^{2}+\mathrm{i}}=w. Thus, an+i=i(w2+1)⋅wn−1=a_{n}+\mathrm{i}=\mathrm{i}\left(w^{2}+1\right) \cdot w^{n-1}= −iwn-\mathrm{i} w^{n}, hence an=−i−iwna_{n}=-\mathrm{i}-\mathrm{i} w^{n}.

Since w3=1w^{3}=1, we have an−3=an(n⩾1)a_{n-3}=a_{n}(n \geqslant 1). This indicates that the sequence {an}\left\{a_{n}\right\} is a periodic sequence with a period of 3. Also, a3=−2ia_{3}=-2 \mathrm{i}, so for all n∈N∗n \in \mathbf{N}^{*}, we have
an2+an+12+an+22=a12+a22+a32=−w4−w2−4=−(w+w2)−4=−3,anan−1+an+1an+2+an+2an=a1a2+a2a3+a3a1=i2w3−2i2w−2i2w2=−1+2w+2w2=−3, \begin{array}{l} a_{n}^{2}+a_{n+1}^{2}+a_{n+2}^{2}=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=-w^{4}-w^{2}-4=-\left(w+w^{2}\right)-4=-3, \\ a_{n} a_{n-1}+a_{n+1} a_{n+2}+a_{n+2} a_{n}=a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1} \\ =\mathrm{i}^{2} w^{3}-2 \mathrm{i}^{2} w-2 \mathrm{i}^{2} w^{2}=-1+2 w+2 w^{2}=-3, \end{array}

Thus, an2+an+12+an+22=anan+1+an+1an+2+an+2ana_{n}^{2}+a_{n+1}^{2}+a_{n+2}^{2}=a_{n} a_{n+1}+a_{n+1} a_{n+2}+a_{n+2} a_{n}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.