Is the number meant to be $3 \underbrace{99\ldots9}_{2025} 6\underbrace{0\ldots0}_{2025}1$? I don't believe the number as stated is a perfect square.

MS_Kekas wrote: Prove that the number \[ 3 \underbrace{99\ldots9}_{2025} \underbrace{60\ldots01}_{2025} \]is a square of a positive integer. $N_1=3 \underbrace{99\ldots9}_{2025} \underbrace{60\ldots01}_{2025}=4.10^{4050}-10^{2025}+6.10^{2024}+1$ $=(2.10^{2025}-1)^2+36.10^{2024}$ And so $\left(2.10^{2025}\right)^2>N_1>\left(2.10^{2025}-1\right)^2$ is not a perfect square aidan0626 wrote: Is the number meant to be $3 \underbrace{99\ldots9}_{2025} 6\underbrace{0\ldots0}_{2025}1$? I don't believe the number as stated is a perfect square. $N_2=3 \underbrace{99\ldots9}_{2025} 6\underbrace{0\ldots0}_{2025}1$ $=4.10^{4052}-10^{2027}+6.10^{2026}+1$ $=4.10^{4052}-4.10^{2026}+1$ And so $N_2=\left(2.10^{2026}-1\right)^2$ is indeed a perfect square

Using the number $3 \underbrace{99\ldots9}_{2025} 6\underbrace{0\ldots0}_{2025}1$ Note that the number has a total of $4053$ digits. Therefore, it is equal to \[3 \cdot 10^{4052} + (10^{2025} - 1) \cdot 10^{2027} + 6 \cdot 10^{2026} + 1 ,\]which can be simplified into \[4\cdot 10^{4052} - 4 \cdot 10^{2026} + 1 = (2 \cdot 10^{2026} - 1)^2 , \]so it is a perfect square.