Running time analysis on computing the largest factor of an integer using Euler's subtraction-based algorithm For the following algorithm,

$X\leftarrow \{(i, n - i) | i = 1, ..., n- 1\}$
while $\max_{(a, b)\in X}b &gt; 0$ do
$\quad X \leftarrow \{(|a - b|, \min\{a, b\})|(a, b) \in X\}$
return $\max_{(a, b)\in X}a$

show that the algorithm does not run in $o(n \log ^k n)$ for any constant $k$.
Summary: the algorithm uses the Euler's subtraction-based algorithm to compute the largest factor of $n$ in $1$ to $n - 1$.
My work: I can prove the upper bound case that $T(n) \in O(n^2)$. Since $n = (a + b)$ is strictly decreasing, gcd computation for each tuple takes at most $n$ operations. Since there are $n-1$ elements, $T(n) \in O(n^2)$.
I can also prove the lower bound case that $T(n) \in \Omega(n \log n)$. The best case for computing gcd of a single tuple occurs if $a = c_1 \cdot b$ or $b = c_2 \cdot a$ for $c_1, c_2 \in \mathbb{N}$. Then, the running time is $\Omega{\log n}$. Therefore, the total running time is $T(n) \in \Omega(n \log n)$.
My problem: But I don't know how to proceed from here. Basically, I need to prove $T(n) \in \Omega{(n^{1+\epsilon})}$ for $\epsilon &gt; 0$. This means I need to prove something polynomial with the gcd computation of a single tuple.
Can anyone help me out with this problem?
It seems the algorithm tries to calculate n-1 gcd's simultaneously, and performs n-1 operations as long as one of the gcd's is not found. And since at least one gcd (gcd (n, 1)) takes n iterations, it is $\Theta (n^2)$. 

Of course that is just due to the stupid implementation, but probably a good example for estimating the runtime of an algorithm, not the runtime of a problem.

You need to show that $n^2$ is not $o(n log^k n)$ for any k.