Dutch Mathematical Olympiad

a. Suppose that $n$ is a self-squared $\ell$-code. Then the last $\ell$ digits of $n^2 - n = n(n - 1)$ are all zeros. This means that $n(n - 1)$ is divisible by $10^\ell$. Vice versa, $n$ is self-squared if $n^2 - n$ is divisible by $10^\ell$ and thus ends in $\ell$ zeros.

b. The number $n(n - 1)$ is divisible by 10 and so, in particular, it is divisible by 5. This means that one of the numbers $n$ and $n - 1$ must be divisible by 5. In other words, the last digit of either $n$ or $n - 1$ must be a 0 or 5. So the only possible final digits for $n$ are 0, 1, 5 and 6.
If the last digit of $n$ is 0, then $n - 1$ is not divisible by 2 or 5. Since $n(n - 1)$ is divisible by $10^\ell = 2^\ell \cdot 5^\ell$ and there can be no factors 2 or 5 in $n - 1$, $n$ must be divisible by $10^\ell$. But that means that the last $\ell$ digits of $n$, which are all the digits of $n$, must all be zeros. Then we get $n = 0$, which is smaller than 2.
If the last digit of $n$ is a 1, then $n$ is not divisible by 2 or 5. Since $n(n - 1)$ is divisible by $10^\ell = 2^\ell \cdot 5^\ell$ and there can be no factors 2 or 5 in $n$, $n - 1$ must be divisible by $10^\ell$. But that means that the last $\ell$ digits of $n - 1$, which are all the digits of $n - 1$, must all be zeros. Then we get $n = 1$, which is smaller than 2.
So all self-quadratic codes $n \ge 2$ end in a 5 or 6.

c. Suppose that $m = c \cdot 10^\ell + n$ is an $(\ell + 1)$-code obtained by putting $c$ in front of $n$. Because of (a), $m$ is self-squared if and only if
$$m(m-1) = (c \cdot 10^{\ell} + n)(c \cdot 10^{\ell} + n-1) = c^2 \cdot 10^{2\ell} + c \cdot 10^{\ell}(2n-1) + n(n-1)$$
is divisible by $10^{\ell+1}$. As $\ell \ge 1$, the term $c^2 \cdot 10^{2\ell}$ is also divisible by $10^{\ell+1}$. The number $n(n-1)$ is divisible by $10^\ell$, because of part (a), and can therefore be written as $d \cdot 10^\ell$. It then follows that
$$c \cdot 10^{\ell}(2n - 1) + d \cdot 10^{\ell} = 10^{\ell} \cdot (c(2n - 1) + d)$$
must be divisible by $10^{\ell+1}$, or in other words that $c(2n - 1) + d$ must be divisible by 10.
Because of part (b), $n$ must end in a 5 or 6. If $n$ ends in a 5, then $2n-1$ is a multiple of ten minus one, i.e. it is of the shape $10k-1$, and we see that $c(10k-1)+d$ must be divisible by 10 and so $-c+d$ must be divisible by 10. On the other hand, if $n$ ends in a 6, then $2n-1$ is a multiple of ten plus one, say $10k+1$, and we see that $c(10k+1)+d$ must be divisible by 10 and so $c+d$ must be divisible by 10.
In both cases, there is a unique possibility for $c$ from the numbers 0, 1, 2, ..., 9 for which $-c+d$ (or $c+d$) is divisible by 10, since, of the ten consecutive numbers $-0+d, -1+d, ..., -9+d$ (or $0+d, 1+d, ..., 9+d$), exactly one is divisible by 10. Therefore, there is a unique digit $c$ that we can put before the $\ell$-code $n$ such that $m$ is a self-squared $(\ell+1)$-code.

d. Suppose that $n \ge 2$ is a self-squared $\ell$-code. Then we will show that $k = 10^\ell + 1 - n$ is also a self-squared $\ell$-code. Because $n \ge 2$, we have $k \le 10^\ell - 1$ and $k$ can thus really be written using $\ell$ digits. From part (a) it follows that $k$ is self-squared if and only if
$$k(k-1) = (10^{\ell} + 1 - n)(10^{\ell} - n) = 10^{2\ell} + 10^{\ell}(1 - 2n) + (n-1)n$$
is divisible by $10^\ell$. The first two terms are always divisible by $10^\ell$ and $(n-1)n$ is divisible by $10^\ell$ because $n$ is self-squared. So it follows that $k$ is self-squared.
Since $n \le 10^\ell - 1$, we have $k \ge 2$. Furthermore, $10^\ell + 1$ is odd, and so one out of $n$ and $k$ is even and the other number is odd. In particular, $n$ and $k$ are thus not equal to each other. Since there are exactly two self-squared $\ell$-codes $\ge 2$, the other self-squared $\ell$-code $m$ must be equal to $k$. It then holds that $m+n=k+n=10^\ell+1$.