Prove or disprove that $8c+1$ is square number. 
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.


Yes, that's true! Here we go: Write $b=a-p$. Then, if you write the expression as a quadratic equation in $c$, you get $3c^2-(2a-p)c+ap-a^2=0.$ Now, by the quadratic formula, see that since we want $c$ to be integer, we must have $(2a-p)^2-12(ap-a^2)=x^2$ for some, say positive, integer $x$. Rewriting in a better way, you get $(4a-2p)^2=x^2+3p^2$, which turns to be $(4a-2p-x)(4a-2p+x)=3p^2.$ Now, since $p$ is prime, there are a few different cases we should look at:
1) $4a-2p-x=1, 4a-2p+x=3p^2$: Isolating $2a-p$ and $x$, you get $x=\frac{3p^2-1}{2}, 2a-p=\frac{3p^2+1}{4}.$ Plugging in these two in quadratic formula, you get $c=\frac{9p^2-1}{24}$ or $c=\frac{-3p^2+3}{24}$, where first one cannot be an integer since numerator is not divisible by $3$, and the latter is not positive. So, there is no solution from this case.
2) $4a-2p-x=3, 4a-2p+x=p^2$. Similarly, you get $2a-p=\frac{p^2+3}{4}, x=\frac{p^2-3}{2}$. Plug in these, and eliminate the possible negative root, you'll get $c=\frac{p^2-1}{8}$, which tells that $8c+1=p^2.$
3) $4a-2p-x=p, 4a-2p+x=3p$. From here, you get $x=p, 2a-p=\frac{p}{2}$, which forces $p$ to be $2$, so you get $a=\frac{3}{2}$, hence no solution from this case too.
Note that these are all possible cases, since $4a-2p-x<4a-2p+x.$
