For every $k\ge 1$, exists $10^{k-1}\le n<10^k$ such that $10^k|n^2-n$?

Question

Show that for every integer $k\ge 1$, exists an integer $n$ with $10^{k-1}\le n<10^k$ such that $10^k|n^2-n$.

In other words, for every $k\ge1$, exists an integer $n$ with $k$ digits such that the last $k$ digits of $n^2$ is the same of the $n$.

Answer 1

5 and 25 show that this is true for k=1 and 2. It also seems true that the units digit is always 5.

It looks like we just have to add a high-order digit, so let's see is that works.

Suppose $n$ works for $k$, so that $10^k | (n^2-n) =n(n-1)$, or $n^2-n = m10^k$.

If the new high-order digit is $d$, then the new number is $n_1 = 10^{k}d+n$ and $10^{k+1} | (n_1^2-n_1)$.

Let $n_1^2-n_1=m_110^{k+1}$. This becomes

$\begin{array}\\ m_110^{k+1} &=n_1^2-n_1\\ &=(10^{k}d+n)^2-(10^{k}d+n)\\ &=10^{2k}d^2+2nd 10^{k}+n^2-10^{k}d-n)\\ &=10^{2k}d^2+d10^{k}(2n-1)+n^2-n\\ &=10^{k}(10^{k}d^2+d(2n-1))+m10^k\\ &=10^{k}(10^{k}d^2+d(2n-1)+m)\\ \end{array} $

Therefore, if we can choose $d$ so that if $10 | (10^{k}d^2+d(2n-1)+m))$, we can do the induction step.

This is the same as $10 | (d(2n-1)+m))$.

When $k=1$, $n=5$ so that $10m=25-5 = 20$ so $m = 2$. The induction step requires $10 |(9d+2)$, so that $d = 2$ works and the next number is $25$.

Every odd number except 5 has an inverse mod 10: the inverses of 1, 3, 7, 9 are 1, 7, 3, 9.

Since $n \equiv 5 \bmod 10$, $2n-1 \not\equiv 5 \bmod 10$, so that $2n-1$ has an inverse mod 10, so we can solve $d(2n-1)+m \equiv 0 \bmod 10 $, so the induction step can proceed.

There is probably a more direct way to do this. Maybe write $n = 10u+5$ with $m10^k = (10u+5)(10u+4) $ so that $5^k | 10u+5$ and $2^k | 10u+4$.

But this works.