Observe that 3025 = (30 + 25)^{2}.

Can you find another four digit number, with all digits distinct, that has this property?

Let our target number be written abcd in base 10. All we know is that:

abcd = (ab + cd)^{2}

First, note that any square number written in base 10 must end with 0, 1, 4, 5, 6, or 9. So these are the possible values for d.

Next, consider the restrictions that these values impose on b.
Using the fact that (b + d)^{2} must end with d, we can
enumerate the possible values for b, given each value of d:

d | b |

0 | 0 |

1 | 0, 8 |

4 | 4, 8 |

5 | 0 |

6 | 0, 8 |

9 | 4, 8 |

Hence b must equal 0, 4 or 8, and the possible combinations, excluding the cases where b = d, are:

b | d |

0 | 1, 5, 6 |

4 | 9 |

8 | 1, 4, 6, 9 |

Now we turn our attention to the other two digits. For each value of a, we can look at the range for abcd, compute the range for ab + cd, and deduce the possible values for c, keeping in mind the fact that all digits are distinct, and a is not zero. The results are summarized in the table below:

a | abcd | ab + cd | c |

1 | 1000-2000 | 32-44 | 2, 3 |

2 | 2000-3000 | 45-54 | 1, 3 |

3 | 3000-4000 | 55-63 | 1, 2 |

4 | 4000-5000 | 64-70 | 1, 2 |

5 | 5000-6000 | 71-77 | 1, 2 |

6 | 6000-7000 | 78-83 | 1, 2 |

7 | 7000-8000 | 84-89 | 0, 1 |

8 | 8000-9000 | 90-94 | 0, 1 |

9 | 9000-9999 | 95-99 | 0 |

We have now reduced the number of possibilities for b and d to 8, and the number for a and c to 17, which gives us 136 candidates for abcd. We can reduce the number of candidate further by means of two additional observations.

The first observation is that any square leaves a remainder of 0 or 1 when divided by 4, and the remainder when abcd is divided by 4 is the same as the remainder when cd is divided by 4. This excludes combinations like 11, 18 or 23 for cd. So for each combination of ac, only some of the combinations for bd are permitted. This is summarized in the following table, where we have also eliminated combinations with duplicated digits:

ac | bd |

12 | 05, 49, 84, 89 |

13 | 06, 86 |

21 | 06, 86 |

23 | 06, 86 |

31 | 06, 86 |

32 | 01, 05, 49, 81, 84, 89 |

41 | 06, 86 |

42 | 01, 05, 81, 89 |

51 | 06, 86 |

52 | 01, 49, 81, 84, 89 |

61 | none |

62 | 01, 05, 49, 81, 84, 89 |

70 | 49, 81, 84, 89 |

71 | 06, 86 |

80 | 49 |

81 | 06 |

90 | 81, 84 |

We are now down to 47 possible values for abcd. One further observation reduces the list some more. Any square leaves a remainder of 0, 1, 4, or 7 when divided by 9. And the remainder when abcd is divided by 9 is the same values as the remainder when a + b + c + d is divided by 9. So from the table above, we can exclude combinations where a + b + c + d leaves a remainder other than 0, 1, 4 or 7 when divided by 9. This yields:

ac | bd |

12 | 49 |

13 | 06, 86 |

21 | 06 |

23 | 86 |

31 | 06, 86 |

32 | 05, 49, 89 |

41 | none |

42 | none |

51 | none |

52 | none |

62 | 01, 05 |

70 | none |

71 | 86 |

80 | none |

81 | none |

90 | 81 |

We're now down to 14 candidates, which can be verified manually:

abcd | ab + cd | (ab + cd)^{2} |

1429 | 43 | 1849 |

1036 | 46 | 2116 |

1836 | 54 | 2916 |

2016 | 36 | 1296 |

2836 | 64 | 4096 |

3016 | 46 | 2116 |

3816 | 54 | 2916 |

3025 | 55 | 3025 |

3429 | 63 | 3969 |

3829 | 67 | 4489 |

6021 | 81 | 6561 |

6025 | 85 | 7225 |

7816 | 94 | 8836 |

9801 | 99 | 9801 |

And at long last, we find the original example, 3025, and the number we're searching for, 9801. I'm exhausted.