= Solution
For the negative-sign equation, take a nonzero separated solution $\phi=X(x)Y(y)$. <Separation of variables> gives $X''=-\mu X$ and $Y''=-(25-\mu)Y$. The zero <Dirichlet boundary conditions> on $X$ force $\mu=m^2$ with a positive integer $m$ and $X=\sin mx$, up to a constant. Zero boundary values on $Y$ similarly require $25-m^2=n^2$ with a positive integer $n$. A zero or negative separation eigenvalue would give linear or hyperbolic factors, whose two zero endpoint values force the trivial solution.
The only positive integer pairs with $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By the complete <Dirichlet eigensystem of a rectangle>, these span the whole eigenspace, so
$$
\boxed{\phi(x,y)=A\sin3x\sin4y+B\sin4x\sin3y,\qquad A,B\in\mathbb R.}
$$
For the positive-sign equation there is \b[only the zero solution]. One can use the same strictly positive eigenvalues of $-\Delta$, or directly apply the <zero Dirichlet solution of the modified Helmholtz equation>. Multiplying by $\phi$ and using <Green's first identity> gives
$$
25\int_\Omega\phi^2\,dx\,dy=\int_\Omega\phi\Delta\phi\,dx\,dy=-\int_\Omega|\nabla\phi|^2\,dx\,dy,
$$
where $\Omega=(0,\pi)^2$. Both nonnegative integrals must vanish, proving $\boxed{\phi=0}$.
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