Minimal size of $a^2+b^2$ such that $ad-bc=1$

Question

If $c,d$ are two relatively prime positive integers, then we can find integers $a,b$ such that $ad-bc=1$. But $a$ and $b$ are not unique: we can replace $a$ with $a+kc$ and $b$ with $b+kd$ for any integer $k$ and get another solution. So there is a solution $(a,b)$ of minimal distance from the origin, in the sense that $a^2+b^2$ is smallest possible, and in fact that's the solution found via the Euclidean algorithm. My question is: what can we say about the size of $a^2+b^2$ for the minimal solution? Is it true that $a^2+b^2$ is always less than $c^2+d^2$?

Answer 1

Comment: See my comment on this question:' Given $a$ in $\gcd(a,b)$ with $a > b > 0$, how can I find $b$ which give the maximum number of steps($l$) for the Euclidean algorithm?'

I found the size of $a^2+b^2$ depends on the number of steps for the Euclidian algorithm. Look at this example from my previous comment:

$1437\times4277-2048\times 3001=1$

1)) Number of steps: 10

2)) $a^2+b^2=6259273$

3)) $c^2+d^2=27298730$

$\Rightarrow a^2+b^2<c^2+d^2$

For the same $d=4277$ we also have:

$1788\times 4277-2825\times 2707=1$

1)) Number of steps: 10

2)) $a^2+b^2=11156149$

3)) $c^2+d^2=25620578$

4)) $\Rightarrow a^2+b^2<c^2+d^2$

Comparing these two we Find:

$(l=10):a^2+b^2=11156149>6259273; (l=6)$

we predict revers for $c^2+d^2$ and it is true and shown below:

$(l=6): c^2+d^2=27298730> 25620578; (l=10)$

So in both cases $a^2+b^2<c^2+d^2$.