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 a1, a2, ... an are called a Diophantine n-tuple if
aras + 1 is a perfect square whenever r ≠ s\.
Tony and David's solutions were almost identical.
Given that ab=q2 − 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=q2 − 1 then ab + 1 = q2 so
ab + 1 is a perfect square.
Next, for bc+1, we substitute c=a+b+2q and expand:
b(a + b + 2q)+ 1
= ab + b2 + 2qb + 1
= q2 − 1 + b2 + 2qb + 1
= q2 + 2qb + b2
= (q + b)2.
Finally, for ac+1 we have
a(a + b + 2q)+ 1 = a2 + ab + 2aq + 1 and in the same way,
substituting ab = q2 − 1 we get (a+q)2 which is
obviously a perfect square. Q.E.D.