Prove that $10101\ldots01$, where the sequence $"01"$ is repeated $k$ times can't be a perfect square for any value of $k$.
All I managed to prove is that the general formula is $1+\frac {100}{99}(100^k-1)$ but I don't know how to continue.
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5It is enough to write $\dfrac{100^k-1}{99}$, where $k>1$. – Oleg567 Sep 08 '17 at 08:18
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Perhaps that you ment to ask that it can't be a perfect square for any value of $k$. Of course that it can't be a perfect square for all values of $k$ since it isn't a perfect square when $k=1$. – José Carlos Santos Sep 08 '17 at 08:19
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Yes, thank you, edited the question (apologies - I am Italian!) – Harry Petrone Sep 08 '17 at 08:23
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1After multiplying it by $33^2$, we'll get expression $$11\times(100^k-1).$$ Maybe roll around it? – Oleg567 Sep 08 '17 at 08:31
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9Hint: Consider this modulo $8$ – Mark Bennet Sep 08 '17 at 08:31
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I also found the recursive relation: f(0)=1 and f(n)=100*f(n-1)+1 – Harry Petrone Sep 08 '17 at 08:34
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An odd square has to be $\equiv 1 \bmod 8$ because$(2m+1)^2=4m^2+4m+1=4m(m+1)+1$ where $m(m+1)$ is even. Any number rendered in base $10$ is a multiple of $1000=125×8$ plus its last three digits, thus congruent with the last three digits $\bmod 8$. But $101 \equiv 5 \bmod 8$, a contradiction.
Oscar Lanzi
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Suppose the statement is true.
Note that any odd square integer can be written as $8m+1$, where $m$ is a positive integer.(*)
Let, $\dfrac{100^k-1}{99}=8m+1$, where $m$ is a positive integer.
Now, $\dfrac{100^k-1}{99}=8m+1\Rightarrow 100^k=99(8m+1)+1=(99\times8)m+100 \\\Rightarrow (99\times8)m=100(100^{k-1}-1)\Rightarrow m=\dfrac{25}{2}\times\dfrac{100^{k-1}-1}{99}$
Note that, for $k>1$, $\dfrac{100^{k-1}-1}{99}$ is an odd integer.
since $k>1$, $m$ can not be an integer. $\Rightarrow \Leftarrow$
(*) Why an odd square integer can be written of the form $8m+1$?
$\Longrightarrow$ Note that $(2n+1)^2=4n^2+4n+1=4n(n+1)+1=8m+1$, since $\dfrac{n(n+1)}{2}$ is an positive integer.(note that $n$ and $n+1$ are consecutive integers)
MAN-MADE
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Perhaps you should add a footnote to justify why $(100^{k-1}-1)/99$ is an odd integer too... – skyking Sep 08 '17 at 09:28
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Well justification of that is the same as the justification that the numbers are $(100^k-1)/99$... – skyking Sep 08 '17 at 09:35
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@MANMAID: I understand that you assume that $\dfrac{100^k-1}{99}=$Odd square integer, but why odd? – Harry Petrone Sep 08 '17 at 09:35
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@MANMAID Yes, but it should be divisible by $99$ as well to make it an integer. – skyking Sep 08 '17 at 09:36
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I don't say it's complicated, but then the fact that odd squares can be written on the form $8m+1$ isn't complicated either. – skyking Sep 08 '17 at 09:37
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are you sure @skyking see that OP commented why $\dfrac{100^k-1}{99}$ is odd! – MAN-MADE Sep 08 '17 at 09:38
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@HarryPetrone your number $1010\dots 01=\dfrac{100^k-1}{99}$, for some $k$, note that that number ends with $1$, how that could be even even number! – MAN-MADE Sep 08 '17 at 09:42
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@HarryPetrone note $2$ does not divides $100^k-1$, since $2$ divides $100^k$ – MAN-MADE Sep 08 '17 at 09:44
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Simpler: Any positive whole number rendered in base$10$ is congruent to its last three digits $\bmod 8$. $101\equiv 5$. – Oscar Lanzi Sep 08 '17 at 10:01
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Maybe help you :$$101010101...01=\\\overline{0101010101...01}=\\\overline{01}+100\times\overline{01}+100^2\times \overline{01}+100^3\times \overline{01}+...+100^m\times \overline{01}=\\\overline{01}(1+100+100^2+100^3+...+100^m)=\\ \overline{01}(\frac{100^{m+1}-1}{99})=\\(\frac{100^{m+1}-1}{99})=k^2 \\100^{m+1}-1=99k^2\\100^{m+1}=99k^2+1=\\100k^2+1-k^2=\\(10k)^2+(1-k^2) \to\\1-k^2=0 \to k=1 \text{ for k=1 this not true }$$
Khosrotash
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This doesn't in anyway prove that it is true for other values of k. – Win Vineeth Sep 08 '17 at 08:39
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@WinVineeth :you are right , I don't claim that's a proof . I said may be help – Khosrotash Sep 08 '17 at 08:41
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It can be proved by the conclusions of the following 2 points:$$$$ 1. Suppose we take k = 1, In this case, 101 is 1 more than 100 (the nearest perfect square). $$$$ similarly, for any value of k, the nearest perfect square will be 1 less than the number. ex. for k=2, 10101 nearest square -> 10000( $100^2$ ),for k=3, 1010101 -> 1,000,000( $ 1000^2$ ) $$$$ 2. Now Observe the following pattern $$$$ $2^2$ - $1^2$ = 4-1 = 3 $$$$ $3^2$ - $2^2$ = 9-4 =5 $$$$ . $$$$ . $$$$ $(x+1)^2$ - $x^2$ = $x^2$ + 2x +1 - $x^2$ = 2x+1 $$$$ Clearly the difference between square of two consecutive CAN NOT BE 1 for any value of k. Ex. for k =1 the nearest squares are 100 and 121 and difference is 21. for k=2 these are 10000 and 10201 and difference is 201 and so on .
So any for any value of k 10101010....01 can not be perfect square.
Ritter7
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1Your first point is false because for $k=3$ we have $1001^2=1002001 < 1010101$ so the directly preceding perfect square is not $1000000$. – skyking Sep 08 '17 at 09:24
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@skyking In fact, $1004^2 = 1008032064$ is even closer, while $1005^2 = 1010050125$ is the closest, although it's larger. – Arthur Sep 08 '17 at 09:27
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In addition another part of the claim is false $10101-10000 = 101$ which is not $1$. The fact that the difference is $1$ is used in the next part of the "proof". This invalidates the "proof" for any $k>1$. – skyking Sep 08 '17 at 09:33