# How to prove that exists perfect square between the number $n10^n$ and $(n+1)10^n$? [closed]

I tried with mathematics induction, but I couldn't prove.

• Is the 'between' supposed to be inclusive or exclusive? Dec 11, 2016 at 22:15
• It is inclusive.
– Bona
Dec 11, 2016 at 22:18
• You could try to show that $\sqrt{(n+1)10^n} - \sqrt{n10^n}$ is an increasing sequence, which means that there will always be an integer between the two. Dec 11, 2016 at 22:18
• @Arthut (minor detail) you also need to prove that this value is bigger than 1, which is true for n=2. Dec 11, 2016 at 22:29
• Can I prove with math induction?
– Bona
Dec 11, 2016 at 22:31

Hint: suppose not, then there is a largest $m$ with $m^2\lt n\cdot10^n$ and we also have $(m+1)^2\gt (n+1)\cdot 10^n$
What can you then say about $(m+1)^2-m^2$?
The difference between consecutive perfect primes $k^2$ and $(k+1)^2$ is $2k+1$.
The difference between $n10^n$ and $(n+1) 10^n$ is $10^n$. Suppose the highest perfect square before the interval is $n10^n$ itself. Then the difference to the next perfect square will be $\sqrt(n10^n)*2+1 = 2\sqrt(n)*\sqrt(10)^n+1$.
This will be smaller than $10^n$ for all $n>=1$. So there must be at least one perfect square in the said interval.