Suppose $\phi(x, y)=X(x) Y(y)$ is a nontrivial separated solution of the Helmholtz equation $\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}=K \phi$ satisfying the boundary conditions $X(0)=X(1)=0, Y(0)=Y(1)=0$
(a) Show that $\phi(x, y)=C \sin (p x) \sin (q y)$ for some $p, q \in \pi \mathbf{Z}, C \in \mathbf{R}$
(b) Deduce that if there is a nontrivial separated solution then $K \in \pi^{2} \mathbf{Z}$
(c) If $K=-2 \pi^{2},$ deduce that the only solutions have the form $C \sin (\pi x) \sin (\pi y), C \in \mathbf{R}$ and sketch the graph of such a solution.
(d) How many solutions can you find when $K=-5 \pi^{2} ?$ When $K=-50 \pi^{2} ?$
(e) Why is the space of solutions to the Helmholtz equation (for fixed $K$ ) a vector space? What is its dimension?
so I have just completed part (c) and I'm now on part (d).
To fill you in, I have found that $K = -\pi^2 (n^2+m^2)$ for some $n,m \in \mathbb{Z}$ and $p = n\pi, q = m\pi$
now I don't really understand what they mean by "how many solutions", take for instance $K = -5\pi^2$, we get then $5 = n^2 + m^2$, implying that $n = 1, m = 2$ or $n = 2, m = 1$ (as if we get $n = -1, m = -2$ we get the same solution as $n = 1, m = 2$ due to sin being odd).
Now, taking these we get the solutions to be in the form of $\sin(\pi x)\sin(2\pi y) $ or $\sin(2\pi x)\sin(\pi y)$ does this mean we have two solutions? I'm not really sure how this relates to the number of solutions we can find and I certainly don't see how this can be a vector space (for part e)