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As @DavidESpeyer suggests, I think you meant ${}+ a$ in place of ${}- a$. As @KevinCasto says, you are then looking for the order of $\begin{bmatrix} 1 & a \\ 1 & 1 \end{bmatrix}$ as an element of $\operatorname{PGL}_2(\mathbb F_p)$, i.e. (since its eigenvalues are $1 \pm \sqrt a$), the order of $1 \pm \sqrt a$ as an element of $\mathbb F_{p^2}^\times/\mathbb F_p^\times$. Certainly this order divides $p + 1 = \lvert\mathbb F_{p^2}^\times/\mathbb F_p^\times\rvert$; but, as @DavidESpeyer also suggests, it need not equal $p + 1$. Indeed, if $p = 5$ and $a = 2$, then the element has order $3$, since $\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}^3 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. (If you really did mean ${}- a$ instead of ${}+ a$, then the same example works; just pretend I took $a = 3$ instead of $a = 2$.)