Hello, I was trying to solve this problem using descarte circle theorem for my maths report. I looked through the solution but I don't understand the part in the answer, where it says the two solutions are $p_{n+1}, p_{n-1}.$ Can someone explain it for me. Thanks!
Consider this form of the equation $$2(p_1^2 + p_n^2 + p_{n+1}^2 + a^2 ) = ( p_1 + p_n + p_{n+1} - a )^2 \tag{1}$$ and take it back a step by replacing $n\to n-1$: $$2(p_1^2 + p_{n-1}^2 + p_{n}^2 + a^2 ) = ( p_1 + p_{n-1} + p_{n} - a )^2 \tag{2}$$ We see that $(2)$ is identical to $(1)$, except that $p_{n-1}$ replaces $p_{n+1}$. Therefore, $p_{n+1}$ and $p_{n-1}$ are the roots of the monic quadratic equation $$2(p_1^2 + p_{n}^2 + x^2 + a^2 ) = ( p_1 + p_{n} + x - a )^2 \tag{3}$$ (Perhaps this trick is discussed in the "remarks above" or Yoshida's solution.) From there, Vieta's formulas allow us to write the coefficient of $x$ in terms of those roots, which gives this recursion: $$p_{n-1} + p_{n+1} = -2(a-p_1-p_n)$$ Since $p_1$ has evidently been assigned the value of $2a$ in the quoted solution, we get the recursion in the stated form. $\square$