Avoidance of patterns $\underbrace{0... 0 ...0}_{k^{*}\text{ times}}1$ of last digits in squares

Question

I would like to propose a following conjecture, which I believe is true, but do not have an idea on how to approach to a possible soulution.

If we take a set of natural squares $1,4,9,16,25,...$ then I believe that last $k$ digits (written in, for example, decimal notation) avoid some patterns, more specifically, I think that a following is true:

There exists some natural number $k^{*}$ such that there is no natural square that ends in $\underbrace{0... 0 ...0}_{k^{*}\text{ times}}1$.

Is this true?

Answer 1

Not true.

Note that $101^2 = 10201$, $1001^2 = 1002001$, $10001^2 = 100020001$, and so on.

In fact , for any square number $k^2$, we have : $$ (l \times 10^n + k)^2 = 10^{n}(l^2 10^n + 2lk) + k^2 $$

This demonstrates two things :

• For all $k^2$, there exist squares ending with $k^2$, $0k^2,00k^2$, etc.

• For each $k^2$ and fixed number of zeros $m$, by varying $l$, we can also find infinitely many square numbers ending in exactly $...\underbrace{0000}_{m\ \mathrm{ zeros}} k^2$.

---

Answer 2

It is not true. For any $k^*$, there exists a number whose square ends like that. In particular,

$$\Big(1\underbrace{00\dots0}_{k^*}1\Big)^2=1\underbrace{00\dots0}_{k^*}2\underbrace{00\dots0}_{k^*}1$$