Solve : $ab(a+b)(a-b)=c^2-1$ As we know that $ab(a+b)(a-b)=c^2$ has no integer solution in $Z^+$.However, it seems that $$ab(a+b)(a-b)=c^2-1$$
has infinite  positive integer solutions,could you prove it?
Here are some of them:
$(a,b,c)=(3, 1, 5), (5, 1, 11), (7, 3, 29), (7, 5, 29), (8, 2, 31), (8, 7,   29), (9, 8, 35), (13, 3, 79), (15, 8, 139), (15, 11, 131), (17, 7,   169), (20, 6, 209), (20, 14, 239), (21, 5, 209), (27, 8, 379), (28,   2, 209), (29, 16, 521)...$
Thanks in advance!
 A: One trick is to reduce it to a Pell-like equation. Given,
$$ab(a+b)(a-b)=c^2-1\tag{1}$$
then (1) is satisfied if,
$$\begin{aligned}a &= p+5q\\
b &= 2q\\
c &= -1+4pq+28q^2\end{aligned}$$
and,
$$p^2-41q^2 = -4$$
One integer solution is $p,q = 64, 10$, and an infinite more. I trust you can take it from here.
A: Let us want
$$a(a-b) = c+1 \text{ and }b(a+b) = c-1$$
For this, we need
$$2 = a(a-b) - b(a+b) = a^2 - 2ab -b^2 = (a-b)^2 - 2b^2$$
Hence, all we want is to solve the Pell's equation
$$x^2-2y^2 =2$$
One trial solution to the above is $(2,1)$ and now generate the rest of $x$ and $y$. From this we get $b=y$ and $a = x+y$. This gives us $$c = \dfrac{a^2+b^2}2$$

Hence, to summarize the solution is as follows. Find the solutions to the Pell's equation
$$2x^2-b^2 = 1$$
(We have replaced $x$ by $2x$ and $y$ by $b$. Hence we get this Pell's equation, which is even more easier to obtain infinite solutions for $x$ and $b$).
We then have
\begin{align}
a & = 2x+b\\
c & = 2x^2 + 2xb + b^2
\end{align}
