Peculiar numbers

If you just look at the units digit, you see that magical numbers must end in $0, 1, 5, $ or $6$. We will show that the only magical numbers that end in $1$ are $1, 01, 001, \dots$.

If $u$ is a two-digit number that ends in 1, then $u = 10x + 1$ for some digit $x$.

\begin{align} u(u - 1) &\equiv 0 \pmod{100}\\ (10n + 1)(10n) &\equiv 0 \pmod{100}\\ 10n &\equiv 0 \pmod{100}\\ n &\equiv 0 \pmod{10}\\ \end{align}

So $u = 01$. Suppose, inductively, that the only $q$-digit magical number that ends in $1$ is $000\dots01$, where there are $q - 1$ $0$s to the left of $1$. Then a $(q+1)$-digit magical number that ends in $1$ must look like $u = 10^q x + 1$ for some digit $x$. Reasoning similar to that shown above will show that $x = 0$. It follows by induction that then only magical numbers that end in $1$ are $1, 01, 001, \dots$.

Similar reasoning will show that the only magical nubers that end in $0$ are $0, 00, 000, \dots$

There is a peculiar pattern with the magic numbers that end in $5$. The first three are listed below.

5²   =     25
25²  =    625
625² = 390625

Note the magical numbers are formed by prepending a digit to the previous magic number.

Note also that the digit prepended is the first nonrepeated digit in $n^2$.

So the next three numbers must be

0625²   =       390625
90625²  =   8212890625
890625² = 793212890625

Where we have to accept $0625$ as a four-digit number.

We can prove this.

Let $N$ be a q-digit magical number that ends in $5$.

We seek a digit $\alpha$ such that the $(q+1)$-digit number $N' = 10^q\alpha + N$ is the next magical number that ends in 5.

\begin{align} N'^2 - N' &\equiv 0 \pod{10^{q+1}}\\ (10^q\alpha + N)^2 - (10^q\alpha + N) &\equiv 0 \pod{10^{q+1}}\\ 2 \cdot 10^q\alpha N - 10^q\alpha + N^2 - N &\equiv 0 \pod{10^{q+1}}\\ -10^q\alpha + N^2 - N &\equiv 0 \pod{10^{q+1}}\\ 10^q\alpha &\equiv N^2 - N \pod{10^{q+1}}\\ \alpha &\equiv \dfrac{N^2 - N}{10^q} \pod{10}\\ \end{align}

Which means that $\alpha$ is the first nonrepeating digit of $N^2$.

To find the magical numbers that end in $6$, note that

6²   =     36  and  5 + 6 = 11
76²  =   5776  and  25 + 76 = 101
376² = 141376  and  625 + 376 = 1001

It seems that the two $q$-digit magical numbers, the one that ends in $5$ and the other that ends in $6$, sum to $10^q + 1$.

It turns out that this is easy to prove.

Let $N$ be a q-digit magical number that ends in $5$. Then

\begin{align} (10^q + 1 - N)^2 - (10^q + 1 - N) &\equiv (1 - N)^2 - (1 - N) \pod{10^q}\\ &\equiv N^2 - N \pod{10^q}\\ &\equiv 0 \pod{10^q}\\ \end{align}

so we find

10001   -   0625 =   9376  and  9376²   =    87909376
100001  -  90625 =  09376  and  09376²  =    87909376
1000001 - 890625 = 109376  and  109376² = 11963109376