Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.
Cubes are of the form \(4m\), \(4m+1\) or \(4m+3\).
Idea. Every integer is either even or odd. Even numbers are multiples of 2, and odd numbers are of the form \(4t+1\) or \(4t+3\). We examine these cases and rewrite the cube as \(4\times(\text{integer})\) plus a small remainder.
Case 1: \(n\) is even.
Let \(n = 2k\).
Then \(n^3 = (2k)^3\).
So \(n^3 = 8k^3\).
Hence \(n^3 = 4\big(2k^3\big)\).
This is of the form \(4m\).
Case 2: \(n\) is odd of the form \(4t+1\).
Let \(n = 4t + 1\).
Compute \(n^3 = (4t+1)^3\).
Expand: \((4t+1)^3 = 64t^3 + 48t^2 + 12t + 1\).
Group multiples of 4: \(64t^3 = 4\cdot 16t^3\), \(48t^2 = 4\cdot 12t^2\), \(12t = 4\cdot 3t\).
Thus \(n^3 = 4\big(16t^3 + 12t^2 + 3t\big) + 1\).
This is of the form \(4m + 1\).
Case 3: \(n\) is odd of the form \(4t+3\).
Let \(n = 4t + 3\).
Compute \(n^3 = (4t+3)^3\).
Expand: \((4t+3)^3 = 64t^3 + 144t^2 + 108t + 27\).
Group multiples of 4: \(64t^3 = 4\cdot 16t^3\), \(144t^2 = 4\cdot 36t^2\), \(108t = 4\cdot 27t\).
Also write \(27 = 24 + 3 = 4\cdot 6 + 3\).
Therefore \(n^3 = 4\big(16t^3 + 36t^2 + 27t + 6\big) + 3\).
This is of the form \(4m + 3\).
Conclusion. For any positive integer \(n\), the cube \(n^3\) is congruent to \(0\), \(1\), or \(3\) modulo \(4\). Equivalently, \(n^3\) has the form \(4m\), \(4m+1\), or \(4m+3\) for some integer \(m\).