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 3 of 4: Apply the two closing equations

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In plain words

Closing the recurrence around the five-cycle restricts the characteristic value and the two initial parameters.

(y^2+y-1)\bigl((y-1)t-s\bigr)=0,\quad (y^2+y-1)\bigl((y-1)s-(y^2-y-1)t\bigr)=0

Detailed analysis

Substituting the expressions from Step 1 into the two remaining equations gives the displayed pair. If $y^2+y-1\ne0$ , the first equation gives $s=(y-1)t$ ; substituting this into the second gives $(y-2)t=0$ . Thus either $y=2$ or $s=t=0$ . If $y^2+y-1=0$ , both closing equations vanish for arbitrary $s,t$ .