Squares Ending in Repeated Digits

Theorem
A square number $n^2$ can end in a repeated digit either:


 * $(1): \quad n^2$ is a multiple of $100$, in which case $n$ is a multiple of $10$
 * $(2): \quad n^2$ ends in $44$ and $n$ ends in $12, 38, 62$ or $88$.

Proof
Let $n \in \Z_{>0}$ end in $a b$.

By the Basis Representation Theorem, $n$ can be expressed as:
 * $n = 100 k + 10 a + b$

for some $k \in \Z_{>0}$ and for $0 \le a < 10, 0 \le b < 10$.

Then:

Thus the nature of the last $2$ digits of $n^2$ are not dependent upon $k$.

So we can ignore all digits of $n$ except the last $2$.

Note that if $b = 0$ we have that $b^2 = 0$ and $20 a b = 0$.

So the last $2$ digits of the square of a multiple of $10$ are both $0$, and $n^2$ is a multiple of $100$.

Let $a = 5 + c$ where $0 \le c < 5$.

We have that:

So the square of a number ending in $c b$, where $0 \le c < 5$, ends in the same $2$ digits as the square a number ending in $a b$ where $a = c + 5$.

It remains to list the integers from $1$ to $49$, generating their squares and investigating their last $2$ digits:

It is seen that the only $n^2$ ending in a repeated digit end in $44$.

The corresponding $n$ are seen to be $12$ and $38$.

Adding $5$ to the $10$s digit of each gives us $62$ and $88$ as further such:

The result follows.