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:
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.