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
$$100 \cdot 2 \cdot \frac{10^{n-1}-1}{9} + 25 = (10k + 5)^2, \quad (1)$$
or equivalently
$$9k^2 + 9k - 2(10^{n-1} - 1) = 0. \quad (2)$$
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:
$$8 \cdot 10^{n-1} = x^2 - 1 = (x - 1)(x + 1) = 4m(m + 1) \Rightarrow$$
$$2^n \cdot 5^{n-1} = m(m + 1), \quad (3)$$
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.