Shortlist JBMO, 2008

Solution:
a) Let $\overline{a b c}=m^{2}$, $\overline{d e f}=n^{2}$ and $\overline{a b c d e f}=p^{2}$, where $11 \leq m \leq 31, 11 \leq n \leq 31$ are natural numbers. So, $p^{2}=1000 \cdot m^{2}+n^{2}$. But $1000=30^{2}+10^{2}=18^{2}+26^{2}$. We obtain the following relations
$$\begin{gathered} p^{2}=\left(30^{2}+10^{2}\right) \cdot m^{2}+n^{2}=\left(18^{2}+26^{2}\right) \cdot m^{2}+n^{2}= \\ =(30 m)^{2}+(10 m)^{2}+n^{2}=(18 m)^{2}+(26 m)^{2}+n^{2} \end{gathered}$$
The assertion a) is proved.

b) We write the equality $p^{2}=1000 \cdot m^{2}+n^{2}$ in the equivalent form $(p+n)(p-n)=1000 \cdot m^{2}$, where $349 \leq p \leq 979$. If $1000 \cdot m^{2}=p_{1} \cdot p_{2}$, such that $p+n=p_{1}$ and $p-n=p_{2}$, then $p_{1}$ and $p_{2}$ are even natural numbers with $p_{1}>p_{2} \geq 318$ and $22 \leq p_{1}-p_{2} \leq 62$. For $m=15$ we obtain $p_{1}=500, p_{2}=450$. So, $n=25$ and $p=475$. We have
$$225625=475^{2}=450^{2}+150^{2}+25^{2}=270^{2}+390^{2}+25^{2}$$