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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.
Diophantine N-tuples
Age 14 to 16
Challenge Level 
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.
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