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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.

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    $\begingroup$ 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. $\endgroup$ – lokodiz Jan 25 '18 at 14:13
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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 $n\ge4$.

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