Question

Disprove the statement that every positive integer is the sum of at most two squares and a cube of nonnegative integers.

Short answer

The statement is not true for \(22\).

Step-by-step solution

Describe the given information

Every positive integer is the sum of at most two squares and a cube of nonnegative integers.

Disprove the given statement

At most \(2\) squares mean no squares, \(1\) square or \(2\) squares.

Let choose the integer \(22\), the squares that this number can obtain are \(1\), \(4\), \(9\), \(16\).

First case no squares:

Number \(22\) is not the cube of an integer.

Second case squares:

16+6=22(6 is not the cube of an integer)

9+13=22(13 is not the cube of an integer)

4+18=22(18 is not the cube of an integer)

1+22=22(21 is not the cube of an integer)

Third case squares:

16+9=25>22 (Not Possible)

16+4+2=22(2 is not the cube of an integer)

16+1+5=22(5 is not the cube of an integer)

9+4+9=22(9 is not the cube of an integer)

And,

9+1+12=22(12 is not the cube of an integer)

4+1+17=22(17 is not the cube of an integer)

Therefore, the statement is not true for 22 and thus it disproven the given statement.