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

Hint: To solve this question, we will first assume a variable a and then by using the Euclid Algorithm, we will write a as 4 using q and r as quotient and remainder respectively. Now as r varies from \[0\le r\le 4,\] so we will consider the cases on r and cube a. \[\Rightarrow {{a}^{3}}={{\left( 4q+r \right)}^{3}}\]
We will use the formula \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\] on RHS and finally generalize the cases of r.

Complete step by step answer:
Let us first assume a number ‘a’ be any arbitrary positive integer. Now we will use the Euclid Division Algorithm to solve further. It is stated as: “Two numbers a and b are given then to compute their HCF, we will use the Euclid Division Algorithm.” Any number a and b can be written as \[a=b{{q}_{1}}+{{r}_{1}}\] where \[{{r}_{1}}\] is the remainder and \[{{q}_{1}}\] is the quotient, \[0\le {{r}_{1}}Again we can write \[b={{r}_{1}}\left( {{q}_{2}} \right)+{{r}_{2}}\] where \[0\le {{r}_{2}}<{{q}_{2}}.\]
And again we can write, \[{{r}_{1}}={{r}_{2}}\left( {{q}_{3}} \right)+{{r}_{3}}.\]
This process goes on until we obtain the remainder as 0. So, we will apply this Euclid Division Algorithm on the positive integer a and 4. A non-negative integer q and r is such that \[a=4q+r;0\le r<4.\]
Cubing both the sides, we get,
\[{{a}^{3}}={{\left( 4q+r \right)}^{3}}\]
Now using the formula \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\] on the RHS of the above equation, we get,
\[{{a}^{3}}={{\left( 4q \right)}^{3}}+{{r}^{3}}+3\left( 4q \right)\left( r \right)\left( 4q+r \right)\]
\[\Rightarrow {{a}^{3}}={{\left( 4q \right)}^{3}}+{{r}^{3}}+12qr\left( 4q+r \right)\]
\[\Rightarrow {{a}^{3}}=64{{q}^{3}}+{{r}^{3}}+12q{{r}^{2}}+48{{q}^{2}}r\]
\[\Rightarrow {{a}^{3}}=\left( 64{{q}^{3}}+48{{q}^{2}}r+12q{{r}^{2}} \right)+{{r}^{2}}......\left( i \right)\]
where \[0\le r<4.\]
There can be several cases on r as r varies from 0 to 4.
Case 1: r = 0
When r = 0, (i) becomes,
\[\Rightarrow {{a}^{3}}=\left( 64{{q}^{3}}+48{{q}^{2}}r+12q{{r}^{2}} \right)+{{r}^{2}}\]
\[\Rightarrow {{a}^{3}}=64{{q}^{3}}\]
\[\Rightarrow {{a}^{3}}=4\times 16{{q}^{3}}\]
Therefore the arbitrary number \[{{a}^{3}}\] is of the form \[4\times 16{{q}^{3}}.\] Therefore, \[{{a}^{3}}\] is of the form \[{{a}^{3}}=4m\] where \[m=16{{q}^{3}}.......\left( i \right)\]
Case 2: r = 1
When r = 1, putting r = 1 in (i), we get,
\[\Rightarrow {{a}^{3}}=64{{q}^{3}}+48{{q}^{2}}+12q+1\]
\[\Rightarrow {{a}^{3}}=4\left( 16{{q}^{3}}+12{{q}^{2}}+3q \right)+1\]
\[\Rightarrow {{a}^{3}}=4m+1\]
\[{{a}^{3}}\] is of the form 4m + 1 where \[m=16{{q}^{3}}+12{{q}^{2}}+3q.\]
Hence, in case II,
\[{{a}^{3}}=4m+1.....\left( iii \right)\]
Case 3: r = 2
When r = 2, putting r = 2 in (i), we have,
\[\Rightarrow {{a}^{3}}=64{{q}^{3}}+144{{q}^{2}}+108q+27\]
\[\Rightarrow {{a}^{3}}=64{{q}^{3}}+144{{q}^{2}}+108q+24+3\]
\[\Rightarrow {{a}^{3}}=4\left( 16{{q}^{3}}+36{{q}^{2}}+27q+6 \right)+3\]
\[{{a}^{3}}\] is of the form 4m + 3 where m is given by \[m=16{{q}^{3}}+36{{q}^{2}}+27q+6\] is an integer…….(iv)
Therefore from equation (ii), (iii) and (iv), we have the cube of any positive is of the form 4m, 4m + 1 or 4m + 3 where m is an integer.

Note:
Students might get confused about considering case III as r = 3 as \[0\le r<4.\] But even if you consider r = 3, then also \[{{a}^{3}}\] i.e. cube of a will be in one of the form 4, 4m + 1, 4m + 3. So case, when r = 3 is given, can be rejected and can be considered if needed to get the same result.