Selected Problems from Open Contests

Question

Prove that none of the integers that contain one 2, one 1, and all the rest zeros, can be expressed as a sum of two perfect squares or as a sum of two perfect cubes.

Accepted Answer

All the numbers described in the problem are divisible by

3

(as their sum of digits is divisible by

3

). Note that all perfect squares leave the remainder

0

or

1

when divided by

3

, and therefore, for the sum of the two perfect squares to be divisible by

3

, they both have to be divisible by

3

. Now, as the numbers are both divisible by

3

, their squares are divisible by

9

and thus the sum of the squares is divisible by

9

. However, the number described in the problem is not divisible by

9

, a contradiction.

Note that when a cube number is divided by

3

, it will leave the remainder of either

0

,

1

or

- 1

. Indeed,

(3 k)^{3} = 9 (3 k^{3})

and

(3 k \pm 1)^{3} = 27 k^{3} \pm 27 k^{2} + 9 k \pm 1 = 9 (3 k^{3} \pm 3 k^{2} + k) \pm 1

. The numbers described in the problem give the remainder

3

when divided by

9

(as their sum of digits gives the remainder

3

when divided by

9

), therefore we conclude that it is impossible to express them as sums of two cubes.