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For the negative-sign equation, take a nonzero separated solution $\varphi =X(x)Y(y)$. Separation of variables gives $X^{\prime \prime }=-\mu X$ and $Y^{\prime \prime }=-(25-\mu )Y$. 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, soFor the positive-sign equation there is 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 $\varphi$ and using Green's first identity giveswhere $\Omega =(0,\pi {)}^2$. Both nonnegative integrals must vanish, proving $\boxed{\varphi =0}$.

Seek a separated nonzero solution $u=X(x)Y(y)$. Division by $XY$ on intervals where the factors are nonzero gives $X^{\prime \prime }/X+Y^{\prime \prime }/Y=-25$. Write $X^{\prime \prime }+\lambda X=0$ and $Y^{\prime \prime }+(25-\lambda )Y=0$, with both factors zero at their endpoints $0,\pi$.

For $X$, a nonpositive $\lambda$ gives only the zero solution under the two endpoint conditions: linear functions vanish identically when $\lambda =0$, and the hyperbolic solutions do likewise for $\lambda <0$. For $\lambda >0$, the endpoints force $X=A\sin (mx)$ with positive integer $m$ and $\lambda =m^2$. Applying the same argument to $Y$ gives $Y=B\sin (ny)$ and $25-m^2=n^2$, with positive integer $n$.

The only positive integer pairs satisfying $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By superposition,To establish that this is the most general sufficiently regular solution, expand $u$ in the complete product Fourier sine series on the square. Substitution, or integration by parts against each sine mode, gives $(25-m^2-n^2)c_{mn}=0$ for every coefficient. Hence only the two displayed modes remain; all boundary terms vanish because of the given zero values. This is the eigenvalue-$25$ case of the Dirichlet eigensystem of a rectangle. The zero solution is included when $A=B=0$.