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}}