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.

 A: 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 $$2c=s(s+1)\implies \color{Blue}{8c+1=(2s+1)^2.}$$
A: This is not a full answer, but it might push someone else in the right direction. The following primitive Maple code searches for solutions within the bounds $1 \leq a,b,c \leq m$ for some $m \in \mathbb{N}.$ (And under the additional constraint $b > a$ since the problem is symmetrical in $a$ and $b.$) For silly programming reasons my assumption is that $b - a$ is prime, so switch the $a$'s and $b$'s from the comments around.
m := 200;
for a to m do
for b from a+1 to m do
for c to m do
if `and`(evalb(sqrt((c*(a+b)+a*b)*(1/3)) = c), isprime(b-a)) then print([a, b, c], b-a, sqrt(8*c+1)) end if
end do end do end do;

The output for $m = 200$ is the following:
[1, 6, 3], 5, 5
[3, 10, 6], 7, 7
[10, 21, 15], 11, 11
[15, 28, 21], 13, 13
[28, 45, 36], 17, 17
[36, 55, 45], 19, 19
[55, 78, 66], 23, 23
[91, 120, 105], 29, 29
[105, 136, 120], 31, 31
[153, 190, 171], 37, 37

The noteworthy thing here is that it is always the case that $8c+1 = (b-a)^2.$ Hence presumably to show that $8c+1$ is a square number one should try to show that it's exactly the square of the prime $b-a.$
A: 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 $4a-2p-x<4a-2p+x.$
A: I think you can make it easier.  Parameterization - that is, the solution of Diophantine equations.
$$3c^2=c(a+b)+ab$$
Have this form.
$$c=(p+s)s$$
$$a=(3s-p)s$$
$$b=(p+s)p$$
Then the difference will be equal.
$$a-b=3s^2-ps-p^2-ps=3s^2-2ps-p^2=4s^2-(p+s)^2=$$
$$=(2s-p-s)(2s+p+s)=(s-p)(3s+p)$$
This number will be easiest if  $s=p+1$
Substitute and find the number itself.
$$8c+1==8s(p+s)+1=8(p+1)(2p+1)+1=$$
$$=8(2p^2+3p+1)+1=16p^2+24p+9=(4p+3)^2$$
All. It is a square.
