Numbers Which Are Both Squares and Cubes Must be of Form 7k, 7k+1

Suppose a number    is both a square and a cube. Then we can write  . Only numbers of certain forms can be both a square and a cube. The table shows such numbers modulo 7.

\[m, \; n\]	\[m^2\]	\[n^3\]
0	\[0 \equiv 0 \; (mod \; 7)\]	\[0 \equiv 0 \; (mod \; 7)\]
1	\[1 \equiv 1 \; (mod \; 7)\]	\[1 \equiv 1 \; (mod \; 7)\]
2	\[4 \equiv 4 \; (mod \; 7)\]	\[8 \equiv 1 \; (mod \; 7)\]
3	\[9 \equiv 2 \; (mod \; 7)\]	\[27 \equiv 6 \; (mod \; 7)\]
4	\[16 \equiv 2 \; (mod \; 7)\]	\[64 \equiv 1 \; (mod \; 7)\]
5	\[25 \equiv 4 \; (mod \; 7)\]	\[125 \equiv 6 \; (mod \; 7)\]
6	\[36 \equiv 1 \; (mod \; 7)\]	\[216 \equiv 6 \; (mod \; 7)\]
7	\[49 \equiv 0 \; (mod \; 7)\]	\[343 \equiv 0 \; (mod \; 7)\]
8	\[64 \equiv 1 \; (mod \; 7)\]	\[512 \equiv 1 \; (mod \; 7)\]
9	\[81 \equiv 4 \; (mod \; 7)\]	\[729 \equiv 1 \; (mod \; 7)\]

From the table, only numbers which have remainder 0 or 1 on division by 7 can be both a square and a cube, so such numbers must be of the form    or  .