According to my solution,
$$y(n) = x(n^2) \tag{1}$$
for $y(n,k)$ i.e output for delayed input $x(n-k)$
$$x_2(n) = x(n-k)$$
so $$y_2(n) = x_2(n) = x\big((n-k)^2\big)$$
and for delayed output signal $y_1(n)$, replace $n$ by $n-k$ in equation (1), so we get,
$$y_1(n) = x\big((n-k)^2\big)$$ and therefore system is time invariant
But in the answers to the book in which this question it says the system is time variant. Can anyone point out the mistake in my steps, and give correct way of solving this sum?