Diophantine n-tuples
Can you explain why a sequence of operations always gives you perfect squares?
Age
14 to 16
Challenge level
Take any whole number $q$. Calculate $ q^2 - 1$. Factorize $ q^2 - 1 $ to give two factors $ a$ and $ b$ (not necessarily $ q+1$ and $ q-1$). Put $ c = a + b + 2q $ . Then you will find that $ ab + 1 $ , $ bc + 1 $ and $ ca + 1 $ are all perfect squares.
Prove that this method always gives three perfect squares.
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 $ . The whole subject started with Diophantus of Alexandria who found that the rational numbers $$ {1 \over 16},\ {33\over 16},\ {68\over 16},\ {105\over 16} $$
have this property. (You should check this for yourself). Fermat was the first person to find a Diophantine 4-tuple with whole numbers, namely $1$, $3$, $8$ and $120$. Even now no Diophantine 5-tuple with whole numbers is known.
have this property. (You should check this for yourself). Fermat was the first person to find a Diophantine 4-tuple with whole numbers, namely $1$, $3$, $8$ and $120$. Even now no Diophantine 5-tuple with whole numbers is known.
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.