40th Hellenic Mathematical Olympiad

Answer: $5^2=25$ and $25^2=225$.

Proof. Since a perfect square cannot have its last digit $2$, $N$ must be of the form: $$N=22\dots 225=22\dots 200+25=100\cdot 2\cdot 11\dots 1+25=100\cdot 2\cdot \frac{10^{n-1}-1}{9}+25,$$ where the digit $2$ there exists $n$ times, $n\ge 1$.

Working mod $10$, then $N=(10k+v{)}^2$, $0\le v<10$, where $k$ is a positive integer, we can see that only in the case $N=(10k+5{)}^2$ results a positive integer leading to $25$. Then we have $$100\cdot 2\cdot \frac{10^{n-1}-1}{9}+25=(10k+5{)}^2,(1)$$ or equivalently $$9k^2+9k-2(10^{n-1}-1)=0.(2)$$ Since $k$ is a positive integer, the discriminant of the equation (2) must be perfect square, that is $\Delta =9(8\cdot 10^{n-1}+1)$ must be perfect square. Thus we must have $$8\cdot 10^{n-1}+1=x^2,$$ for some odd integer $x>1$. If $x=2m+1$, $m\ge 1$, then: $$8\cdot 10^{n-1}=x^2-1=(x-1)(x+1)=4m(m+1)\Rightarrow$$ $$2^n\cdot 5^{n-1}=m(m+1),(3)$$ with $(m,m+1)=1$. We distinguish cases with respect to $n$:

If $n=1$, then $m=1$, $k=0$ and $N=25$. If $n=2$, then $m=4$, $k=1$ and $N=225$. * If $n\ge 3$, since $5^{n-1}>4^{n-1}=2^{2n-2}>2^n$, from equation (3) we get: $$2^n=m\text{ and }{5}^{n-1}=m+1,$$ and so $m=5^{n-1}-1=4\cdot (5^{n-2}+\cdots +1)>4\cdot 4^{n-2}=4^{n-1}>2^n=m$, absurd.