I tried with mathematics induction, but I couldn't prove.
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$\begingroup$ Is the 'between' supposed to be inclusive or exclusive? $\endgroup$– ServaesDec 11, 2016 at 22:15
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$\begingroup$ It is inclusive. $\endgroup$– BonaDec 11, 2016 at 22:18
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2$\begingroup$ 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. $\endgroup$– ArthurDec 11, 2016 at 22:18
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$\begingroup$ @Arthut (minor detail) you also need to prove that this value is bigger than 1, which is true for n=2. $\endgroup$– ypercubeᵀᴹDec 11, 2016 at 22:29
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$\begingroup$ Can I prove with math induction? $\endgroup$– BonaDec 11, 2016 at 22:31
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2 Answers
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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 $(m+1)^2-m^2$?
And take it from there to get a contradiction.
answered Dec 11, 2016 at 22:20
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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.
