Tony (State College Area High School, PA, US) and David (The Lawrenceville School, USA) both cracked this problem.Tony and David's solutions were almost identical.

The numbers $a_1, a_2, ... a_n$ are called a Diophantine n-tuple if $a_ra_s + 1$ is a perfect square whenever $r \neq s$.

Given that $ab=q^2 - 1$, and $c = a + b + 2q$, we must show that $ab + 1$, $bc + 1$, and $ac + 1$ are all perfect squares.

For the first one, as $ab=q^2 - 1$ then $ab + 1= q^2$, so $ab + 1$ is a perfect square.

Next, for $bc+1$, we substitute $c=a+b+2q$ and expand:

$$\eqalign{ b(a + b + 2q)+ 1 &= ab + b^2 + 2qb + 1 \cr &= q^2 - 1 + b^2 + 2qb + 1 \cr &= q^2 + 2qb + b^2 \cr &= (q + b)^2. } $$

Finally, for $ac+1$, we have $a(a + b + 2q)+ 1 = a^2 + ab + 2aq + 1$ and in the same way, substituting $ab = q^2 - 1$, we get $(a+q)^2$ which is obviously a perfect square. Q.E.D.