Positive real solutions in cyclic system with alternating 4 and 1

[anemone]

Find all positive real solutions of the system below:

$a_1+\dfrac{1}{a_2}=4,\,a_2+\dfrac{1}{a_3}=1,\cdots,a_{99}+\dfrac{1}{a_{100}}=4,\,a_{100}+\dfrac{1}{a_1}=1$

 

[lfdahl]

Spoiler

Every term, $a_i$ can be expressed by $a_1$ as follows:
\[a_k = \frac{(k-1)-(\frac{k}{2}-1)a_1}{2k-(k-1)a_1} \; \; \; \; k = 2,4,..,100. \\\\\\ a_j = \frac{2(j-1)-(j-2)a_1}{j-\left ( \frac{j-1}{2} \right )a_1}\; \; \; j =1,3,5,..,99. \\\\\\ a_{100}=\frac{99-49a_1}{200-99a_1}\; \; \; \; \; \;and\;\;\;\;\; a_{100}+\frac{1}{a_1}=1\Rightarrow a_1^2-4a_1+4 = 0\]
The positive (and only) solution is: $a_1 = 2$

So
\[a_k = \frac{1}{2}\; \; \; k = 2,4,..,100 \\\\ a_j = 2\; \; \; \; j = 1,3,..,99\]

 

[anemone]

Thanks for participating, lfdahl! :) I think it really is a great idea to relate $a_{100}$ with $a_1$ again and with the two well defined formulas, how to solve further would then be as clear as daylight. Well done!:)

Another method that I saw online that I would like to share:

By AM-GM:

$a_1+\dfrac{1}{a_2}\ge2\sqrt{\dfrac{a_1}{a_2}},\,\cdots,\,a_{100}+\dfrac{1}{a_1}\ge2\sqrt{\dfrac{a_{100}}{a_1}}$

Multiplying we get

$\left( a_1+\dfrac{1}{a_2} \right)\left( a_2+\dfrac{1}{a_3} \right)\cdots\left( a_{100}+\dfrac{1}{a_1} \right)\ge2^{100}$,

From the system of equations we get

$\left( a_1+\dfrac{1}{a_2} \right)\left( a_2+\dfrac{1}{a_3} \right)\cdots\left( a_{100}+\dfrac{1}{a_1} \right)=2^{100}$,

so all thoseinequalities are equalities, i.e.

$a_1+\dfrac{1}{a_2}=2\sqrt{\dfrac{a_1}{a_2}}$

$\left(\sqrt{a_1}-\dfrac{1}{\sqrt{a_2}} \right)^2=0,\,\,\,\rightarrow a_1=\dfrac{1}{a_2}$

and analogously $a_2=\dfrac{1}{a_3},\,\cdots,\,a_{100}=\dfrac{1}{a_1}$.

Hence, we get $a_1=2,\,a_2=\dfrac{1}{2},\,\cdots\,,a_{99}=2,\,a_{100}=\dfrac{1}{2}$.