### Square remainders

N | N^{2} | rem |
N | N^{2} | rem |

1 | 1 | 1 | 7 | 49 | 1 |

2 | 4 | 4 | 8 | 64 | 4 |

3 | 9 | 9 | 9 | 81 | 9 |

4 | 16 | 4 | 10 | 100 | 4 |

5 | 25 | 1 | 11 | 121 | 1 |

6 | 36 | 0 | 12 | 144 | 0 |

**"All squares, on division by twelve, leave a remainder which is also a
square."*** (Old Algebra book)*
**rem** = remainder mod twelve

The statement is also true, by the way, if we replace the word
"twelve" by "five", "sixteen", or "twenty-four".

Can you find other solutions?

### Quick Squares

(NB "10" means any number base)
**A handy rule for squaring numbers**
(mental, or pencil & paper)

1) Find the difference, "d" between the number, "n", and the next multiple of "10" (above or below n).

2) If you needed to add the difference, d, to get the multiple, also form the
number n-d;

if you needed to subtract the difference, also form the number n+d.

3) Multiply these numbers (n+d) and (n-d) and then add the square of
the difference, d^{2}, to this answer.
Which says that: n^{2} = (n-d)(n+d) + d^{2}

This may sound rather complicated, so here are some examples in
numbers:

a) (base ten)

n = "38"; next multiple of 10 is 38+2, i.e. 40; this is n+d, so d = 2

also form n-d: 38 - 2 = 36.

Square of 38 is 36 x 40 + 2^{2} = 1440+4 = 1444

b) (base seven)
n = "36" ; next multiple is 40, or 36 + 1

also form 36 - 1 = 35.

Square of 36 is 35 x 40 + 1^{2} = 2060 + 1 = 2061.

c) (any base greater than 4)

n = "12" ; nearest multiple is 10, or 12 - 2;

also form 12 + 2 = 14.

Square of 12 is 14 x 10 + 2^{2} = 140 + 4 = 144

And the proof? [ (n-d)(n+d) ] +d^{2} =[ n^{2} -nd +nd + d^{2} ] + d^{2}

= [n^{2} - d^{2}] + d^{2} = n^{2}.