2022/064) How to solve the Diophantine equation ab+a+b+82=z2, where (a,b,z) are positive integers & g.c.d(a,b)=1?

We have $ab + a + b + 82 = z^2$

or $ab + a + b + 1 + 81 = z^2$ we do so that we can facrot LHS

or $(a+1)(b+1) = z^2 - 81 = (z-9)(z+9)$ 

we can choose a +  1 = z - 9 and b+1 = z + 9 (this shall not give all solutions but some 

so a = z - 10 and b = z + 8

they have to be co-prime so both are odd say z = 2 m + 1

so a = 2m - 9 and b = 2m + 9

for these to be co-prime there should not a factor of 18 (that is the difference)

so 2m - 9 shall not have a factor 3 or m must not have a factor 3

so a = 2m - 9 and b = 3m + 9 m > 5 and of the forn $3k \pm 1$