# Prove that there are infinitely many perfect squares starting with 2018

Prove that there are infinitely many perfect squares starting with 2018(as for example 20187049,201810436 are perfect square.

First of all I had assumed that there are finitely many perfect square numbers starting with 2018 and thought proving this by contradiction. Now let M is the maximum number starting with 2018. I can realize that we can find another number N\gt M which is a perfect square starting with 2018 . But I failed to find any direct method of solving this. Somebody please give any hint or the full solution.

N.B. If you have any types of confusion please command I will surely answer. So command before down voting or closing this question.

• Professor Vector's answer is nice. However, if you are told that $20187049$ is a perfect square, then so is $20187049 \times 100^n$ for any positive integer $n$, and you're done. – lokodiz Jan 25 '18 at 14:13

Since $$\sqrt{2019\cdot10^n}-\sqrt{2018\cdot10^n}=\frac{10^n}{\sqrt{2019\cdot10^n}+\sqrt{2018\cdot10^n}}>\frac{10^n}{2\sqrt{2019\cdot10^n}}=\sqrt{\frac{10^n}{4\cdot2019}}>1$$ for $n\ge4$, there must be an integer in that interval. And that means there is an integer square with $n+4$ decimals starting with $2018$ for every $n\ge4$.