= Solution
Let $M=\mathbb R^d/\Lambda$ be a <flat torus>, where $\Lambda$ is a full-rank <Euclidean lattice>, and use the nonnegative <Laplace-Beltrami operator> $\Delta=-\sum_{k=1}^d\partial_k^2$. The <dual lattice> is
$$
\Lambda^*=\{w\in\mathbb R^d:\langle w,v\rangle\in\mathbb Z\text{ for every }v\in\Lambda\}.
$$
For $w\in\Lambda^*$, the function $\phi_w(x)=\operatorname{vol}(M)^{-1/2}e^{2\pi i\langle w,x\rangle}$ is periodic under $\Lambda$, so it defines a function on $M$, and direct differentiation gives $\Delta\phi_w=4\pi^2|w|^2\phi_w$. Integrating characters on a fundamental parallelepiped proves that these functions are orthonormal. After choosing a lattice basis, the ordinary <Fourier series> theorem on $\mathbb R^d/\mathbb Z^d$ proves completeness, so there are no additional <eigenvalues>. Thus the <spectrum of a flat torus> is
$$
\boxed{\operatorname{Spec}(\Delta_M)=\{4\pi^2|w|^2:w\in\Lambda^*\},}
$$
with multiplicity equal to the number of dual vectors of the given length; the zero <eigenvalue> occurs once.
It remains to show that a two-dimensional <Euclidean lattice> is determined up to an orthogonal transformation by its vector-length multiset. This is the <two-dimensional lattice reconstruction from vector lengths>. Let $v$ be a shortest nonzero vector, of length $a$. It is primitive: $v=ku$ with integer $|k|>1$ would give a shorter nonzero vector. The lattice vectors on its line are therefore precisely the integer multiples of $v$. Subtract two occurrences of each length $ka$, $k=1,2,\ldots$, from the length multiset. This removes exactly one primitive lattice line, even when other vectors happen to have the same lengths. The smallest remaining length $b$ is the length of a shortest vector $w$ independent of $v$.
These two vectors form a <basis> of the lattice. Otherwise a nonzero coset of $\mathbb Zv+\mathbb Zw$ has a lattice representative $u=sv+tw$ with $|s|,|t|\leq1/2$. It cannot have $t=0$, since primitivity of $v$ would then force $u=0$. It is therefore independent of $v$, while
$$
|u|\leq|s|a+|t|b\leq\tfrac12(a+b)\leq b.
$$
The inequality is strict: if both coefficients are nonzero, the vectors are not parallel and the triangle inequality is strict; if $s=0$, then $|u|\leq b/2$. This contradicts the choice of $w$.
The <covolume> $A$ is also determined by the length multiset. If $N(R)$ counts lattice vectors in a radius-$R$ disk, then
$$
N(R)=\frac{\pi R^2}{A}+O(R),\qquad A=\lim_{R\to\infty}\frac{\pi R^2}{N(R)}.
$$
To justify the estimate, tile the plane by a bounded fundamental parallelogram. Only cells meeting a fixed-width neighbourhood of the disk boundary contribute an error; their total number is $O(R)$. The lattice basis $v,w$ has <Gram matrix>
$$
\begin{pmatrix}a^2&c\\c&b^2\end{pmatrix},\qquad a^2b^2-c^2=A^2.
$$
Thus $|c|=\sqrt{a^2b^2-A^2}$ is determined. Replacing $w$ by $-w$ makes $c\geq0$, so the <Gram matrix> is determined entirely. Equal <Gram matrices> give an orthogonal map between the lattice bases and hence between the lattices.
Two isospectral two-dimensional <flat tori> consequently have orthogonally equivalent <dual lattices>. Taking duals shows that their original lattices are orthogonally equivalent too, and the orthogonal map descends to an <isometry> of the quotient tori. \b[Two isospectral flat tori of dimension two are isometric.]
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