Clock Squares

Age 11 to 14 Challenge Level:

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 $1^2, 2^2, \ldots, m^2$ form a cycle of length $m$ which is repeated again and again.

Each cycle ends in a zero because $m^2 = 0$ (mod $m$).

The sequence of square numbers starts with $1, 4, 9, \ldots, ({1\over 2}(m - 1))^2$ and then these numbers are repeated in reverse order ending with $\ldots, 9, 4, 1$ 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 $1^2 = (m - 1)^2$, $2^2 = (m - 2)^2$ etc. and the two 'middle' terms $ ({1\over 2}(m - 1))^2 = ({1\over 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 $0 \leq r \leq m - 1$.