TITLE: Prove or disprove that $8c+1$ is square number.
QUESTION [11 upvotes]: Let $a,b,c$ be positive integers, with $a-b$ prime, and $$3c^2=c(a+b)+ab.$$  Prove or disprove that $8c+1$ is square number.

REPLY [4 votes]: Let $a,b,c\in\mathbb{Z}^+$, with $a−b\in\mathbb{P}$, and 
$$4c^2=(a+c)(b+c).$$ Then $$a=r^2t-c \,\,\,\,\,\,\text{and}\,\,\,\,\,\,b=s^2t-c\,\,\,\,\,\,\text{and}\,\,\,\,\,\ 2c=rst$$ for some $r, s, t:=\gcd(a+c,b+c) \in\mathbb{Z}^+.$ Since $$a-b=t(r-s)(r+s)$$ is prime, we must have $r=s+1$ and $t=1$. Hence $$2c=s(s+1)\implies \color{Blue}{8c+1=(2s+1)^2.}$$