C1. Determine all triples $(a, b, c)$ of positive integers with the following properties:

- it holds that $a<b<c$ and $a, b$ and $c$ are three consecutive odd numbers;
- the number $a^{2}+b^{2}+c^{2}$ consists of four identical digits.
C1. Since $a, b$ and $c$ are consecutive positive odd numbers, we can write: $a=2 n-1, b=2 n+1$ and $c=2 n+3$, with $n$ a positive integer.

$\mathrm{Now}$ we calculate:

$$
\begin{aligned}
a^{2}+b^{2}+c^{2} & =(2 n-1)^{2}+(2 n+1)^{2}+(2 n+3)^{2} \\
& =\left(4 n^{2}-4 n+1\right)+\left(4 n^{2}+4 n+1\right)+\left(4 n^{2}+12 n+9\right) \\
& =12 n^{2}+12 n+11
\end{aligned}
$$

This must be equal to a number consisting of four times the digit $p$. The number $12 n^{2}+12 n$ thus consists of four digits, the first two of which are $p$ and the last two are $p-1$. Since $12 n^{2}+12 n$ is divisible by 2, $p-1$ must be even. This leaves the possibilities 1100, 3322, 5544, 7766, and 9988 for $12 n^{2}+12 n$. However, it must also be divisible by 3, which leaves only 5544.

We have now found that $12 n^{2}+12 n=5544$, so $n^{2}+n=\frac{5544}{12}=462$. This can be rewritten as $n^{2}+n-462=0$. This is a quadratic equation, which we can factorize: $(n-21)(n+22)=0$. Since $n$ must be a positive integer, the only solution is $n=21$. From this, we calculate the only triplet that satisfies the condition: $(a, b, c)=(41,43,45)$.