Theorem: Squares Ending in Repeated Digits

A square number $n^2$ can end in a repeated digit {{iff}} 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:
{{begin-eqn}}
{{eqn | l = n^2
      | r = \paren {100k + 10 a + b}^2
      | c = 
}}
{{eqn | r = 100^2 k^2 + 2000 k a + 200 k b + 100 b^2 + 20 a b + b^2
      | c = 
}}
{{eqn | o = \equiv
      | r = 20 a b + b^2
      | rr = \pmod {100}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \paren {10 a + b}^2
      | r = \paren {10 \paren {5 + c} + b}^2
      | c = 
}}
{{eqn | r = \paren {10 c + 50 + b}^2
      | c = 
}}
{{eqn | r = 100 c^2 + 2 \times 500 c + 2 \times 10 b c + 2 \times 50 b + b^2
      | c = 
}}
{{eqn | r = 100 c^2 + 1000 c + 100 b + 20 b c + b^2
      | c = 
}}
{{eqn | r = 1000 c + 100 b + \paren {10 c + b}^2
      | c = 
}}
{{eqn | o = \equiv
      | r = \paren {10 c + b}^2
      | rr = \pmod {100}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = 01^2
      | r = 01
}}
{{eqn | l = 02^2
      | r = 04
}}
{{eqn | l = 03^2
      | r = 09
}}
{{eqn | l = 04^2
      | r = 16
}}
{{eqn | l = 05^2
      | r = 25
}}
{{eqn | l = 06^2
      | r = 36
}}
{{eqn | l = 07^2
      | r = 49
}}
{{eqn | l = 08^2
      | r = 64
}}
{{eqn | l = 09^2
      | r = 81
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 11^2
      | r = 121
}}
{{eqn | l = 12^2
      | r = 144
      | c = $n^2$ ends in $44$ and $n$ ends in $12$
}}
{{eqn | l = 13^2
      | r = 169
}}
{{eqn | l = 14^2
      | r = 196
}}
{{eqn | l = 15^2
      | r = 225
}}
{{eqn | l = 16^2
      | r = 256
}}
{{eqn | l = 17^2
      | r = 289
}}
{{eqn | l = 18^2
      | r = 324
}}
{{eqn | l = 19^2
      | r = 361
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 21^2
      | r = 441
}}
{{eqn | l = 22^2
      | r = 484
}}
{{eqn | l = 23^2
      | r = 529
}}
{{eqn | l = 24^2
      | r = 576
}}
{{eqn | l = 25^2
      | r = 225
}}
{{eqn | l = 26^2
      | r = 676
}}
{{eqn | l = 27^2
      | r = 729
}}
{{eqn | l = 28^2
      | r = 784
}}
{{eqn | l = 29^2
      | r = 841
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 31^2
      | r = 961
}}
{{eqn | l = 32^2
      | r = 1024
}}
{{eqn | l = 33^2
      | r = 1089
}}
{{eqn | l = 34^2
      | r = 1156
}}
{{eqn | l = 35^2
      | r = 1225
}}
{{eqn | l = 36^2
      | r = 1296
}}
{{eqn | l = 37^2
      | r = 1369
}}
{{eqn | l = 38^2
      | r = 1444
      | c = $n^2$ ends in $44$ and $n$ ends in $38$
}}
{{eqn | l = 39^2
      | r = 1521
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 41^2
      | r = 1681
}}
{{eqn | l = 42^2
      | r = 1764
}}
{{eqn | l = 43^2
      | r = 1849
}}
{{eqn | l = 44^2
      | r = 1936
}}
{{eqn | l = 45^2
      | r = 2025
}}
{{eqn | l = 46^2
      | r = 2116
}}
{{eqn | l = 47^2
      | r = 2209
}}
{{eqn | l = 48^2
      | r = 2304
}}
{{eqn | l = 49^2
      | r = 2401
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = 62^2
      | r = 3844
}}
{{eqn | l = 88^2
      | r = 7744
}}
{{end-eqn}}
The result follows.
{{qed}}