= Solution
{c}
Write $d=-c>0$. Suppose, for a contradiction, that only finitely many sign changes occur. Since every $a_n\ne0$, the sign is then constant for $n\geq N$. Set $b_n=|a_n|>0$ on that tail. Multiplying the <recurrence relation> by this constant sign gives
$$
b_{n+1}=\frac12\left(b_n-\frac d{b_n}\right)>0,
\qquad 0<b_{n+1}<\frac12b_n.
$$
Consequently $b_{N+j}<2^{-j}b_N$ for $j\geq1$, and $b_n\to0$ by comparison with a <geometric sequence>. But positivity of $b_{n+1}$ requires $b_n^2>d$ at every step, which is incompatible with $b_n\to0$. Thus the <negative-parameter square-root iteration changes sign infinitely often>.
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