I have an idea on how to implement this in proof talk but I'm not quite there.
If $n = 10^x$, then $n^2= 10^{2x}$ which has $2x+1$ digits but I don't know where to go from there.
Any help?
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@ThomasAndrews typo sorry, I meant surjective – Tingo Hugo Jan 12 '23 at 21:51
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$(10^n)^2$ has $2n+1$ digits, and $(4\cdot 10^n)^{2}$ has $2n+2$ digits. – Thomas Andrews Jan 12 '23 at 21:56
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How does $(4⋅10^n)^2$ have 2n+2 digits – Tingo Hugo Jan 12 '23 at 22:00
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From this post, the number of digits of a positive integer $n$ written in base $10$ is $\lfloor \log_{10} n\rfloor +1$. – Scene Jan 12 '23 at 22:00
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I'm not doing the whole thing for you. Think about it for a moment. @TingoHugo – Thomas Andrews Jan 12 '23 at 22:16
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You just need to provide an $n$ for each $m\in \Bbb N$ such that $c(n) = m$. You've already got half of them: if $m = 2k+1$ is odd, then $n= 10^k$ works.
The other half is to just add or remove exactly one digit. Well, if you cut $10^k$ in half, then squaring will cut $10^{2k}$ in quarter, which will have one less digit*. So, if $m=2k$ is even, then $n= 10^k /2$ works.
*because we would need to cut by a factor of more than $10$ to reduce the number of digits by more than $1$
Alex Jones
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1In fact, $10^k-1$ is another possible choice for „one digit less“ – Hagen von Eitzen Jan 12 '23 at 22:09
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I don't fully follow how $10^k/2$ does that though, like when m = 2k how come $10^k/2$ accounts for the even digits? – Tingo Hugo Jan 12 '23 at 22:12
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$10^k /2$ squares to $10^{2k} / 4 = 2.5\cdot 10^{2k-1}$ which has $2k$ digits. In general, $a \cdot 10^b$ has $b+1$ digits when $a\in [1,10)$. – Alex Jones Jan 12 '23 at 22:15
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If $n$ is odd, then there surely is a square with $n$ digits, namely $10^{n-1}$.
For even $n>0$, take $10^{n/2}-1$ and square it. Now prove that $$ 10^{n-1}\le 10^n-2\cdot10^{n/2}+1<10^n $$
egreg
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If $m\ge3$ and $m^2$ has $n$ digits, then $m^2<(m+1)^2=m^2+2m+1<m^2+3m\le2\cdot m^2$ so that $(m+1)^2$ has either $n$ or $n+1$ digits. Then the result follows because any sequence of natural numbers that begins with $1$, is unbounded, and grows by at most $1$ per step, is quite clearly surjective.
This method works also for the number of digits of cubes or fourth powers. It even works for the number of digits of 2023rd powers if we replace “surjective” with “surjective up to finely many exceptions”.
Hagen von Eitzen
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One of $m$ or $m+1$ is even, so either $10^m$ or $10^{m+1}$ is a square. Because $3 \lt \sqrt{10}$, there must be an integer $k$ such that $10^m \leq k^2 \lt 10^{m+1}=10 \cdot 10^m$. If $10^{m+1}=k^2$, then $c \left (\frac k3 \right)=m$
Robert Shore
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