Problem 2. Determine all numbers of the form

20252025 ... 2025

(consisting of one or more consecutive blocks of 2025 ) that are perfect squares of positive integers.

Solution 1Solution 2Solution 3

Solution 1

Expanding this number gives

$$(10^{4k}+10^{4k-4}+...+10^4+1)\times 2025$$

is a perfect square.
Thus,

$$(10^{4k}+10^{4k-4}+...+10^4+1)$$

is a perfect square.
We thus want

$$10^{4(k+1)}-1=9999y^2 \rightarrow 10^{4(k+1)}-9999y^2=1.$$

Substituting $x=10^{2(k+1)}$ reveals this is a Pell's equation. We prove that there are no solutions where $x$ is a power of $10$.
Clearly, $x=100, y=1$ is a solution.
The other solutions of the Pell's equation are given by

$$x_n+y_n\sqrt{9999}=(100+\sqrt{9999})^n.$$

Expansion of $x_n$ gives that the maximum power of $10$ dividing $x_n$ is at most $2$, so it cannot be a perfect power of $10$.

Solution 2

Here is an elementary approach : The problem reduces to finding all $n\ge 0$ such that

$1+10^4+\ldots+\left(10^4\right)^n=\frac{10^{4(n+1)}-1}{10^4-1}$

is a perfect square. Factoring out the $9$ from the denominator, we need to find when

$10^{4(n+1)}-1=1111\times x^2=11\times 101\times x^2$,

or in other words, when

$\left(10^{n+1}-1\right)\left(10^{n+1}+1\right)\left(10^{2(n+1)}+1\right)=11\times 101\times x^2$.

Call these factors $A, B$ and $C$ respectively. Note that $A, B$ and $C$ are pairwise coprime. Therefore, since $11$ and $101$ are primes, one of $A$, $B$ or $C$ must be a square itself (the one which doesn't contain the $11$ nor the $101$). Clearly $C$ cannot be a square, $B$ cannot either (because it is $2$ mod $3$), and $10^{n+1}-1$ cannot be a square either when $n\ge 1$ because it would be $-1$ mod $4$.

Therefore it remains to check the case $n=0$, which gives us a valid solution.

Solution 3
Since $2025$ is a square, we want $10001000\ldots 10001$ to be a square. We claim this works if it is only $1$. The main expression is $10^0 + 10^4 + \cdots + 10^{4(k-1)} = \frac{10^{4k}-1}{9999}$. So we work with $(10^k - 1)(10^k + 1)(10^{2k} + 1) = 11 \cdot 101 \cdot m^2$ for some $m$. All three factors are pairwise coprime, so one of them must be a square, since $11$ and $101$ are primes. However, $10^k - 1$ is not a square for $k\geq 2$ by mod 4, $10^k + 1$ is not a square by mod 3 and $10^{2k}+1$ is not a square, since no two positive integer squares have difference $1$.
Solution 4 Solution 4