Prove that $10101\ldots01$ can't be a perfect square. 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.
 A: 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.
A: 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)
A: Maybe help you :$$101010101...01=\\\overline{0101010101...01}=\\\overline{01}+100\times\overline{01}+100^2\times \overline{01}+100^3\times \overline{01}+...+100^m\times \overline{01}=\\\overline{01}(1+100+100^2+100^3+...+100^m)=\\
\overline{01}(\frac{100^{m+1}-1}{99})=\\(\frac{100^{m+1}-1}{99})=k^2 \\100^{m+1}-1=99k^2\\100^{m+1}=99k^2+1=\\100k^2+1-k^2=\\(10k)^2+(1-k^2) \to\\1-k^2=0 \to k=1 \text{ for k=1 this not true }$$
A: 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.
