Good answers to this popular question came from Georgina Baxter,
Suzanne Dash, Emily Savell, Helen Almand, Naomi Pearson, Catherine
Renwick and Sang-mi Kim, Davison High School, Worthing; from James
Dotti and Shabbir Tejani, Jack Hunt School, Peterborough; and from
Rachel Walker, Christianne Eaves, Suzanne Abbott, Peggy Brett,
Fiona Conroy, Nisha Doshi and Helen Battersby, The Mount School,
York.
| Mod 10 |
Mod 11 |
Mod 13 |
Mod 15 |
Mod 17 |
Mod 19 |
Mod 21 |
| 1 |
1 |
1 |
1 |
1 |
1 |
1 |
| 4 |
4 |
4 |
4 |
4 |
4 |
4 |
| 9 |
9 |
9 |
9 |
9 |
9 |
9 |
| 16 |
5 |
3 |
1 |
16 |
16 |
16 |
| 25 |
3 |
12 |
10 |
8 |
6 |
4 |
| 36 |
3 |
10 |
6 |
2 |
17 |
15 |
| 49 |
5 |
10 |
4 |
15 |
11 |
5 |
| 64 |
9 |
12 |
4 |
13 |
7 |
1 |
| 81 |
4 |
3 |
6 |
13 |
5 |
18 |
| 100 |
1 |
9 |
10 |
15 |
5 |
16 |
| 121 |
m 2 =0 |
4 |
1 |
2 |
7 |
16 |
| 144 |
1 |
1 |
9 |
8 |
11 |
18 |
| 169 |
4 |
m 2 =0 |
4 |
16 |
17 |
1 |
| 196 |
9 |
1 |
1 |
9 |
6 |
5 |
| 225 |
5 |
4 |
m 2 =0 |
4 |
16 |
15 |
| 256 |
3 |
9 |
1 |
1 |
9 |
6 |
| 289 |
3 |
3 |
4 |
m 2 =0 |
4 |
16 |
| 324 |
5 |
12 |
9 |
1 |
1 |
9 |
| 361 |
9 |
10 |
1 |
4 |
m 2 =0 |
4 |
| 400 |
4 |
10 |
10 |
9 |
1 |
1 |
For odd values of m, the square numbers 12, 22, …, m2 form a
cycle of length m which is repeated again and again.
Each cycle ends in a zero because m2 = 0 (mod m).
The sequence of square numbers starts with 1, 4, 9, …, (1/2(m −1))2 and then these numbers are repeated in reverse order ending with
…, 9, 4, 1 followed by 0. The whole sequence is then repeated.
Apart from the zeros there are an even number of terms in each cycle
arranged symmetrically. The first (m−1) terms form a 'symmetric' pattern
with 12 = (m − 1)2, 22 = (m − 2)2 etc. and the two 'middle' terms
(1/2(m − 1))2 = (1/2(m + 1))2 have a difference equal to
the modulus.
You might like to prove algebraically that the patterns are symmetric
(you only need to work out the difference of two squares) and also that the
cycles will repeat because (m + r)2 = (km + r)2 (mod m) where
k is any whole number and 0 ≤ r ≤ m − 1.