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.
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.
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.