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Recently on StackOverflow, a question was asked about a way to find numbers which form the trailing digits of their square. Here are some examples of such numbers.

$6 ^ 2 = 36$

$376 ^ 2 = 141 376$

$90625 ^ 2 = 8212890625$

I suggested a solution that would build such numbers from right to left, making it sub-linear. This allowed to test if such numbers exist for large numbers of digits. It would seem these numbers are fairly common. But is there infinitely many of them?

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Yes, there are infinitely many such numbers.

Suppose $n$ is a $d$-digit number with that property, i.e. the last $d$ digits of $n^2$ are the digits of $n$ (in order). That means $n^2 \equiv n \pmod{10^d}$, or equivalently $n\cdot (n-1) \equiv 0 \pmod{10^d}$. Since $n$ and $n-1$ are coprime, we must have one of the following cases:

  1. $n \equiv 0 \pmod{10^d}$,
  2. $n \equiv 1 \pmod{10^d}$,
  3. $n \equiv 0 \pmod{2^d}$ and $n-1 \equiv 0 \pmod{5^d}$,
  4. $n \equiv 0 \pmod{5^d}$ and $n - 1 \equiv 0 \pmod{2^d}$.

In case 1, the only way that $n$ can have at most $d$ digits is if $n = 0$. We can count that or not. In case 2, we must have $n = 1$, otherwise $n$ would have too many digits. We can count this or not, it's again not very interesting.

The other cases, 3 and 4, are interesting. By the Chinese remainder theorem, there are solutions to these systems of congruences, unique modulo $10^d$. In particular, there is a unique $n_d$, $0 \leqslant n_d < 10^d$, with $n_d \equiv 0 \pmod{5^d}$ and $n_d \equiv 1 \pmod{2^d}$, and a unique $m_d$, $0 \leqslant m_d < 10^d$, with $m_d \equiv 0 \pmod{2^d}$ and $m_d \equiv 1 \pmod{5^d}$. It is fairly obvious that we have $m_d,n_d > 0$ in fact.

It is possible that $n_d$ or $m_d$ actually has fewer than $d$ digits - for example $n_4 = 625 = n_3$ has only three digits - but there are infinitely many distinct $n_d$ and infinitely many distinct $m_d$. And apart from $0$ and $1$, these are all numbers forming the trailing digits of their squares.

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    $\begingroup$ I'm not sure how often it happens that $n_d = n_{d+1}$ or $m_d = m_{d+1}$. Gut feeling says it'll be rare, but my gut feeling is known to be sometimes wrong. $\endgroup$ – Daniel Fischer Apr 5 '18 at 20:44
  • $\begingroup$ It would seem it happens fairly often using the script in the question's link. $\endgroup$ – Olivier Melançon Apr 5 '18 at 20:53

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