7.60 Find all natural numbers that cannot be expressed as the difference of squares of any natural numbers.
(49th Moscow Mathematical Olympiad, 1986)
[Solution] For the given natural number $n$, consider the equation
$$
n=x^{2}-y^{2}.
$$

where $x$ and $y$ are natural numbers.
Since $x^{2}-y^{2}=(x+y)(x-y)$, and $x+y$ and $x-y$ have the same parity, $n$ is either divisible by 4 or has the form $4k \pm 1$.

Next, we prove that, except for $n=1$ and $n=4$, any natural number $n$ of the form $4k$ or $4k \pm 1$ can be expressed as $n=x^{2}-y^{2}$.
Indeed, if $n=4k$, then
$$
(x-y)(x+y)=4k
$$

We can take $x=k+1, y=k-1 \quad(k>1)$.
If $n=4k \pm 1$, i.e., $n$ is an odd number $2l+1$, then
$$
(x-y)(x+y)=2l+1.
$$

We can take $x=l+1, y=l \quad(l>0)$.
In summary, only $n=1, n=4$, and natural numbers of the form $4k+2$ cannot be expressed as the difference of squares of two natural numbers.