Problem 4

Find all real solutions

x_1,\ldots,x_5

x_i+x_{i+2}=yx_{i+1}

for

i=1,\ldots,5

, where subscripts are reduced modulo

5

Step 1 of 4: Propagate the recurrence

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In plain words

A second-order recurrence has two degrees of freedom before the cycle closes.

x_1=s,\quad x_2=t,\quad x_3=yt-s,\quad x_4=(y^2-1)t-ys,\quad x_5=(1-y^2)s+(y^3-2y)t

Detailed analysis

Set $x_1=s$ and $x_2=t$ . Applying $x_{i+2}=yx_{i+1}-x_i$ successively gives $x_3=yt-s$ , $x_4=(y^2-1)t-ys$ , and $x_5=(1-y^2)s+(y^3-2y)t$ ; the closing equations will restrict $y,s,t$ .