Question 22

Show that cube of any positive integer is of the form 4m, 4m+1 or 4m+3, for some integer m.

Solution:

To show that the cube of any positive integer is of the form $$4m$$, $$4m+1$$, or $$4m+3$$ for some integer $$m$$, we can follow these steps:

Step 1: Define the Positive Integer
Let $$a$$ be any positive integer. According to the division algorithm, we can express $$a$$ in the form:
$$a=4q+r$$
where $$q$$ is some integer, and $$r$$ is the remainder when $$a$$ is divided by 4. The possible values for $$r$$ are $$0$$, $$1$$, $$2$$, or $$3$$.

Step 2: Calculate the Cube for Each Case of $$r$$
Now we will calculate $$a^3$$ for each possible value of $$r$$:

1. Case 1: $$r=0$$
$$a=4q⟹{\mathrm{a3}}^{}=(4q{)3}^{}=64{\mathrm{q3}}^{}=4(16{\mathrm{q3}}^{})⟹{\mathrm{a3}}^{}\text{is of the form}4m$$

2. Case 2: $$r=1$$
$$a=4q+1⟹{\mathrm{a3}}^{}=(4q+1{)3}^{}=64{\mathrm{q3}}^{}+48{\mathrm{q2}}^{}+12q+1$$
$$=4(16{\mathrm{q3}}^{}+12{\mathrm{q2}}^{}+3q)+1⟹{\mathrm{a3}}^{}\text{is of the form}4m+1$$

3. Case 3: $$r=2$$
$$a=4q+2⟹{\mathrm{a3}}^{}=(4q+2{)3}^{}=64{\mathrm{q3}}^{}+48{\mathrm{q2}}^{}+12q+8$$
$$=4(16{\mathrm{q3}}^{}+12{\mathrm{q2}}^{}+3q+2)⟹{\mathrm{a3}}^{}\text{is of the form}4m+0\text{(which is}4m\text{)}$$

4. Case 4: $$r=3$$
$$a=4q+3⟹{\mathrm{a3}}^{}=(4q+3{)3}^{}=64{\mathrm{q3}}^{}+48{\mathrm{q2}}^{}+36q+27$$
$$=4(16{\mathrm{q3}}^{}+12{\mathrm{q2}}^{}+9q+6)+3⟹{\mathrm{a3}}^{}\text{is of the form}4m+3$$

Step 3: Conclusion
From the calculations above, we see that for any positive integer $$a$$, the cube $${\mathrm{a3}}^{}$$ can be expressed in one of the following forms:
– $$4m$$
– $$4m+1$$
– $$4m+3$$

Thus, we have shown that the cube of any positive integer is of the form $$4m$$, $$4m+1$$, or $$4m+3$$ for some integer $$m$$.