If we have two numbers a and b such that ab + 1 is square, then it’s always possible to find a number c for which ac + 1 and bc + 1 are both square. For example, 8 × 3 + 1 = 25 = 52, and 8 × 21 + 1 = 169 = 132 and 3 × 21 + 1 = 64 = 82.
Proof:
If ab + 1 = m2, then set c = a + b + 2m. Now
ac + 1 = a2 + ab + 2am + 1 = a2 + 2am + m2 = (a + m)2
bc + 1 = ab + b2 + 2bm + 1 = b2 + 2bm + m2 = (b + m)2
Via Edward Barbeau, Power Play, 1997.