It can be shown that a magic square which satisfies these three conditions, must also be pandiagonal. One example is the following most-perfect magic square of order n=8.

1 | 16 | 17 | 32 | 53 | 60 | 37 | 44 |

63 | 50 | 47 | 34 | 11 | 6 | 27 | 22 |

3 | 14 | 19 | 30 | 55 | 58 | 39 | 42 |

61 | 52 | 45 | 36 | 9 | 8 | 25 | 24 |

12 | 5 | 28 | 21 | 64 | 49 | 48 | 33 |

54 | 59 | 38 | 43 | 2 | 15 | 18 | 31 |

10 | 7 | 26 | 23 | 62 | 51 | 46 | 35 |

56 | 57 | 40 | 41 | 4 | 13 | 20 | 29 |

But this doesn't mean on the other side that all pandiagonal magic squares are also most-perfect. For example, the following square is pandiagonal and even more - any four adjacent integers forming a 2x2-subsquare sum to

1 | 16 | 57 | 56 | 17 | 32 | 41 | 40 |

58 | 55 | 2 | 15 | 42 | 39 | 18 | 31 |

8 | 9 | 64 | 49 | 24 | 25 | 48 | 33 |

63 | 50 | 7 | 10 | 47 | 34 | 23 | 26 |

5 | 12 | 61 | 52 | 21 | 28 | 45 | 36 |

62 | 51 | 6 | 11 | 46 | 35 | 22 | 27 |

4 | 13 | 60 | 53 | 20 | 29 | 44 | 37 |

59 | 54 | 3 | 14 | 43 | 38 | 19 | 30 |

But when you sum any two integers along a diagonal distant n/2=4, you won't get T=65. So this is a pandiagonal, but not a most-perfect magic square.