Math Olympiad 1988 Problem 6 – Canonical Solution Without Vieta Jumping

Question

There is a recent question about this famous problem from 1988 on this forum, but I'm unable to respond to this because the subject is closed for me (insufficient reputation).
Therefore this new post on the subject.

here's the link to the earlier subject

The problem:
Let a and b be positive integers. Let $$ k={{a^2+b^2}\over{1+ab}}
$$ Show that if $k$ is an integer then $k$ is a perfect square.

The question was: Is there a more direct and intuitive way to arrive at the solution (instead of the usual proof using Vieta jumping and proof by contradiction)?

After seeing this hilarious Numberphile video on Youtube i decided i had to give it a go myself.

The solution i came up with below seems to me more canonical because it's not a proof by contradiction, and it arrives at the actual solution . It doesn't just prove $k$ is a square but more specifically:

$$ k= {\gcd(a,b)}^2 $$

Here goes the proof:

Let $a > 1$ and $b > a$. In this form: $$ k={{1+{{b^2}\over{a^2}}}\over{{{1}\over{a^2}}+{{b}\over{a}}}} $$ it's easy to see that $$ {b\over a} – 1 < k < {b\over a} + 1 \enspace (*) $$
When $b\over{a}$ is a fraction there are exactly two integers in the interval $$ \left\langle{b\over a} – 1 , {b\over a} + 1\right\rangle $$
(thanks zyx for pointing out my error!).
However when $a$ divides $b$ then $b\over{a}$ becomes an integer.
Then the above open interval can contain only one integer
which of course must be $k$ itself! (**).
We'll use that fact below.

Now if we write $$b=ka+c$$
we see that $$\mid c \mid < a$$ because of (*) above. Substituting this expression in the expression for $k$ gives:
$$ k={{a^2+(ka+c)^2}\over{1+a(ka+c)}}
\enspace or \enspace k(1-ca) = {a^2+c^2} \enspace \enspace
$$
We see that $c$ must be negative and replace $a$ with $b'$ and $-c$ with $a'$ to get:
$$ k={{a'^2+b'^2}\over{1+a'b'}}
$$
Iterating this process is in fact the Euclidean algorithm (Slightly different, but similar. See remark zyx below) for finding the greatest common divisor of $a$ and $b$ eventually stopping at :
$$ a' = \gcd(a,b) \enspace $$
but by that time ${{b'}\over{a'}}$ is no longer a fraction but an integer, so it must be equal to $k$ because of $(**)$. ($a'$ divides $b'$ because $a' = \gcd(a,b)$)
so:
$$ {{b'}\over{a'}}={{a'^2+b'^2}\over{1+a'b'}} \text{ or }
b' =a'^3 \text{ or }k =a'^2 \text{ or }k= {\gcd(a,b)}^2
$$

Is the above correct or did i miss something? If not could this be a more direct way to prove the famous problem 6?

Let me know what you think!

UPDATE 12/9/2016:
see this link for another solution

Answer 1

Geometric solution to Q6: Consider a rectangle with sides 𝑎, 𝑏 Its diagonal has length $$ \sqrt{a^2+b^2} $$

This diagonal is a side of square A

Area A $$= a^2+b^2$$ Area B = $$ 𝑐. \sqrt{a^2+b^2}$$ Geometrically Q6 becomes equation (1) , $$ 𝑘 = Area A / Area B $$

Length 𝑎 is projected onto the side square A to yield a length, 𝑐

Now, $$ 𝑐 = 𝑎/cosϴ $$ And $$cosϴ = 𝑏/√{a^2+b^2}$$ So $$𝑐 = 𝑎/𝑏. √{a^2+b^2}$$ Area B $$= 𝑎/𝑏. {a^2+b^2}$$ Then from (1) $$𝑘 = 𝑏/𝑎 $$ (2)

Now if Area B = $$𝑎𝑏 + 1$$ as in Q6 then $$𝑎𝑏 + 1 = a/b.{a^2+b^2}$$ $$b^2 + 𝑏/𝑎 = {a^2+b^2}$$ $$𝑏 = a^3$$ From (2) $$𝑘 = a^2,$$ a perfect square