Question

Use Euclid’s division lemma to show that the square of any positive integer is either of the form $3m$ or $3m+1$ for some integer m, but not for the form $3m+2$.

Answer 1

Hint: Use the division algorithm which $a=bq+r$ where a is dividend, b is divisor, q is quotient and r is remainder after that take $b=3$. When a number is divided by 3 then the remainder will be either 0, 1, 2 and now using these facts proved to get a result.

Complete step-by-step answer:

In the question we have to show that the square of any positive integer is either in the form of $3m,3m+1$ for some integer m, but not of the form $3m+2$ using Euclid’s division lemma.
If a, b are positive numbers where a id dividend and b is divisor then we can write a relation as,
$a=bq+r$, 0$\le$r$<$b
Where q is quotient and r is remainder.
Here the value of b will be 3. So, if a number is divided by 3 then the remainder will be 0, 1, 2. Therefore, a can be equal to $3q,3q+1$ or $3q+2$.
So, a is either $3q,3q+1$ or $3q+2$. So, we will take one by one case.
Let at first assume that $a=3q$ then
${{a}^{2}}={{\left( 3q \right)}^{2}}=9{{q}^{2}}$
So, ${{a}^{2}}=3{{\left( 3q \right)}^{2}}$
Here, ${{\left( 3q \right)}^{2}}$ can be considered as an integer m. So, ${{a}^{2}}$ can be considered or written as $3m$.
Now let’s assume \[a=3q+1\] then, \[{{a}^{2}}={{\left( 3q+1 \right)}^{2}}\] we will expand using formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. So, we get \[{{a}^{2}}=9{{q}^{2}}+1+6q\] which can be further written as,
${{a}^{2}}=3\left( 3{{q}^{2}}+2q \right)+1$
Now, here we can consider $\left( 3{{q}^{2}}+2q \right)$ as integer m. So, ${{a}^{2}}$ can be considered or written in the form of $3m+1$.
Finally, let’s assume $a=3q+2$ then ${{a}^{2}}={{\left( 3q+2 \right)}^{2}}$, we will expand using formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, we get ${{a}^{2}}=9{{q}^{2}}+6q+4$ which can be written as,
${{a}^{2}}=3\left( 3{{q}^{2}}+2q+1 \right)+1$
Here, we can consider $3{{q}^{2}}+2q+1$ as an integer m. So, ${{a}^{2}}$ can be considered or written in the form of $3m+1$.
So, hence we can say that the square of any integer is in the form of either $3m$ or $3m+1$ but not $3m+2$.
Hence, proved.

Note: In the third case instead of $3q+2$ one can also use $3q-1$ and then proceed with the question which will process the same results as done $3q+2$.