$pk+1$ divides $(p^2-1)/2$ implies $k=1$? 
Let $p$ be an odd prime. Let $k\in\mathbb{Z}^+$ be such that $pk+1$ divides $(p^2-1)/2$. Would this suffice to say $k=1$?

 A: Yes. Assume that $pk+1$ divides $\frac{p^2-1}{2}$. This means that $pk+1 \mid (p^2-1)$. If $m(pk+1)=p^2-1$, then we have:
$$m(pk+1) \equiv p^2-1 \pmod{p} \implies m \equiv p-1 \pmod{p} \implies m \geqslant p-1$$
So 
$$p^2-1=m(pk+1)\geqslant(p-1)(pk+1)\implies p+1\geq pk+1\implies k=1$$
A: Let $(p^2-1)/2=m(pk+1)$ with $m\in \Bbb N.$ Then $p^2-2p(mk)-(2m+1)=0.$ By the Quadratic Formula, $p=mk\pm \sqrt {(mk)^2+(2m+1)}.$
So $(mk)^2 +(2m+1)=s^2$ for some $s\in \Bbb N.$ But if $k>1$ then we have $(mk)^2<s^2<(mk+1)^2,$ which implies $mk<s<mk+1.$ 
A: So we have $$2pk+2\mid p^2-1$$ This forces $$2pk+2\mid 2k(p^2-1)-(2kp+2)p= -2p-2$$
so $$2pk+2\mid p+2\implies 2pk+2\leq 2p+2\implies k\leq 1$$
A: By Euclid $\,\color{#c00}d := \overbrace{(pk\!+\!1,p\!-\!1)=(\color{#90f}{k\!+\!1},p\!-\!1)}^{\textstyle pk\!+\!1\equiv \color{#90f}{k\!+\!1}\ \pmod{p\!\equiv\! 1}\!\!} \color{#c00}{\le k\!+\!1},\,$ so with $\, \overbrace{\color{0a0}{n:= (p+1)/2}}^{\color{#0a0}{\textstyle{2n\!-\!1 = p}}\ \ \ \ \ \ \ \ \ \ }>1$
thus $\ pk\!+\!1\mid (p\!-\!1)\color{#0a0}n\Rightarrow\ pk\!+\!1\mid \color{#c00}dn\le (\color{#c00}{k\!+\!1})n\,\Rightarrow\, (p\!-\!n)k \le \color{#0a0}{\underbrace{n\!-\!1}_{\textstyle p\!-\!n}}\,\Rightarrow\, k\le 1$
