Tony Cardell, age 14, State College Area High School, PA, USA and David Aaronson, age 15, The Lawrenceville School, USA both cracked this problem.

The numbers $a_{1}, a_{2}, . . . a_{n}$ are called a Diophantine n-tuple if $a_{r} a_{s} + 1$ is a perfect square whenever $r \neq s$ \.

Tony and David's solutions were almost identical. Given that $ab = q^{2} - 1$ and $c = a + b + 2 q$ 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:

$\begin{matrix} b (a + b + 2 q) + 1 & = ab + b^{2} + 2 qb + 1 \\ = q^{2} - 1 + b^{2} + 2 qb + 1 \\ = q^{2} + 2 qb + b^{2} \\ = (q + b)^{2} . \end{matrix}$

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