Answer:

(ab)^{2} = (10a + b)^{2} = 100a^{2} + 20ab +
b^{2} = ccdd

Thus, d = b^{2} (mod 10).

**Note**: Mod 10 means the remainder when the number is divided by 10.
Two numbers are equal mod 10 if after you divide them by 10, their remainders
are equal. Or when you subtract them, their difference is a multiple of 10.
15 and 5 are equal mod 10 since 15 - 5 is a multiple of 10.

Single-Digits Squared Table | |||||

a | b | a^{2} | b^{2} | c | d |

0 | 0 | 00 | 00 | 0 | 0 |

1 | 1 | 01 | 01 | 1 | 1 |

2 | 2 | 04 | 04 | 4 | 4 |

3 | 3 | 09 | 09 | 9 | 9 |

4 | 4 | 16 | 16 | 6 | 6 |

5 | 5 | 25 | 25 | 5 | 5 |

6 | 6 | 36 | 36 | 6 | 6 |

7 | 7 | 49 | 49 | 9 | 9 |

8 | 8 | 64 | 64 | 4 | 4 |

9 | 9 | 81 | 81 | 1 | 1 |

Therefore, d = 0, 1, 4, 5, 6 or 9.

**Step I**:

If (ab)^{2} = ccdd, then ccdd = 11 * (0c0d) ⇒ 11 | ccdd ⇒
11 | (ab)^{2} ⇒ 11 | ab (since 11 is prime).

Thus, a = b.

**Note**: Since 11 | ab, then 11^{2} | (ab)^{2} ⇒
11^{2} | ccdd ⇒ 11 | c0d0 ⇒ 11 | (c + d).

**Step II**:

a = b = 0 ⇒ c = d = 0 (single digit).

∴ a = b ≥ 1.

a ≥ 1 ⇒ c ≥ 1 ⇒ cc ≥ 11 ⇒ a ≥ 3 (see silver digits in Squares table).

b ≥ 1 ⇒ d ≥ 1 (see gray digits in Squares table).

**Note**: Since 11 | (c + d), c ≠ 0, d ≠ 0, and c + d ≤ 18, then
c + d = 11.

**Step III**:

For any odd, single-digit number b, b^{2} has an even tens digit.
Thus, 20ab + b^{2} = dd (mod 10) must have the same odd digit for both
places. However, 20ab + b^{2} will always have an even digit in the tens
place because b^{2} is even in the tens place (see
red digits in Squares table) and
20ab is 2ab in the tens place, which is also an even number. Thus, b and d have
to be even. Here are all the possibilities for further elaboration.

- d = 1 ⇒ b = 1 ⇒ 100a
^{2}+ 20a + 1 = cc11 ⇒ 20a + 1 = 11 ⇒ 2a = 1 (mod 10) (impossible) - d = 1 ⇒ b = 9 ⇒ 100a
^{2}+ 180a + 81 = cc11 ⇒ 180a + 81 = 11 ⇒ 8a + 8 = 1 (mod 10) (impossible) - d = 5 ⇒ b = 5 ⇒ 100a
^{2}+ 100a + 25 = cc55 ⇒ 25 = 55 (impossible) - d = 9 ⇒ b = 3 ⇒ 100a
^{2}+ 60a + 9 = cc99 ⇒ 6a = 9 (mod 10) (impossible) - d = 9 ⇒ b = 7 ⇒ 100a
^{2}+ 140a + 49 = cc99 ⇒ 14a + 4 = 9 (mod 10) (impossible)

Thus, d = 4 or 6.

**Step IV**:

By the same reasoning, for any even, single-digit number b,
**if** b^{2} has the first digit odd, 20ab + b^{2} = dd (mod
10) must have the same even digit for both places. However, 20ab + b^{2}
will always have an odd digit in the tens place because b^{2} is odd in
the tens place (see blue digits in
Squares table) and 20ab is 2ab in the tens place which is
even. Thus, d = 6 is impossible.

- d = 6 ⇒ b = 4 ⇒ 100a
^{2}+ 80a + 16 = cc66 ⇒ 8a + 1 = 6 (impossible) - d = 6 ⇒ b = 6 ⇒ 100a
^{2}+ 120a + 36 = cc66 ⇒ 120a + 3 = 6 (impossible)

Hence, d = 4 is our only solution and b = 2 or 8.

**Step V**:

Since b = 2 or 8, a ≥ 3 and a = b, then a = b = 8.

Since d = 4 and c + d = 11, then c = 7.

(ab)^{2} = 88^{2} = 7744 = ccdd.

**Thus, a = b = 8, c = 7 and d = 4 is the unique solution.**