Question

Use Euclid division lemma to show that square of any positive integer cannot be of the form $5m+2,5m+3$ for some integer m.

Answer 1

Hint: We know, Euclid division lemma for any two positive integers a and b is given by $a=bm+r$ where m is any positive integer and r is a number less than b i.e. $0\le r\text{ }<\text{ }b$. For this sum we will assume b as 5 to get $a=5m+r$. Then we will take all possibilities of r to form numbers of all forms. Squaring those numbers will give us the type of numbers squared numbers can be. Then, we will check that no squared number is of the form $5m+2,5m+3$.

Complete step by step answer:
As we know, Euclid's division lemma for any two integers a and b is given by $a=qb+r$ where q is any positive integer and $0\le r\text{ }<\text{ }b$.
Here, let us assume that two positive integers are a and 5. So our numbers are of the form (according to Euclid's lemma): $a=5q+r$ where r = 0, 1, 2, 3, 4.
Now all the numbers can be of the form $a=5q+r$ where r = 0, 1, 2, 3, 4.
So our actual numbers are:
Putting r = 0, we get $a=5q+0$.
$\Rightarrow a=5q\cdots \cdots \cdots \cdots \cdots \left( 1 \right)$
Putting r = 1, we get $a=5q+1$.
$\Rightarrow a=5q+1\cdots \cdots \cdots \cdots \cdots \left( 2 \right)$
Putting r = 2, we get $a=5q+2$.
$\Rightarrow a=5q+2\cdots \cdots \cdots \cdots \cdots \left( 3 \right)$
Putting r = 3, we get $a=5q+3$.
$\Rightarrow a=5q+3\cdots \cdots \cdots \cdots \cdots \left( 4 \right)$
Putting r = 4, we get $a=5q+4$.
$\Rightarrow a=5q+4\cdots \cdots \cdots \cdots \cdots \left( 5 \right)$
So number can be of the form (1), (2), (3), (4) or (5).
Let us now square these numbers to find the possible type of numbers.
From (1), $a=5q$.
Squaring both sides we get: ${{\left( a \right)}^{2}}={{\left( 5q \right)}^{2}}\Rightarrow {{a}^{2}}=25{{q}^{2}}$.
Now 25 can be written as $5\times 5$ so ${{a}^{2}}=5\left( 5{{q}^{2}} \right)=5m$ where $m=5{{q}^{2}}$.
Here the square of the number is of the form 5m where m is any integer.
From (2), $a=5q+1$.
Squaring both sides we get: ${{a}^{2}}={{\left( 5q+1 \right)}^{2}}\Rightarrow {{a}^{2}}=25{{q}^{2}}+1+10q$.
Now taking 5 common from $25{{q}^{2}}$ and 10q we get: ${{a}^{2}}=5\left( 5{{q}^{2}}+2q \right)+1\Rightarrow {{a}^{2}}=5m+1$ where $m=5{{q}^{2}}+2q$.
Here the square of the numbers is of the form $5m+1$.
From (3), $a=5q+2$.
Squaring both sides, we get: ${{a}^{2}}={{\left( 5q+2 \right)}^{2}}\Rightarrow {{a}^{2}}=25{{q}^{2}}+4+20q$.
Now taking 5 common from $25{{q}^{2}}$ and 20q we get: ${{a}^{2}}=5\left( 5{{q}^{2}}+4q \right)+4\Rightarrow {{a}^{2}}=5m+4$ where $m=5{{q}^{2}}+4q$.
Here the square of the number is of the form $5m+4$.
From (4), $a=5q+3$.
Squaring both sides we get: ${{a}^{2}}={{\left( 5q+3 \right)}^{2}}\Rightarrow {{a}^{2}}=25{{q}^{2}}+9+30q$.
Splitting 9 into 5+4 and then taking 5 common from $25{{q}^{2}}$,5,30q we get:
${{a}^{2}}=5\left( 5{{q}^{2}}+6q+1 \right)+4\Rightarrow {{a}^{2}}=5m+4$ where $m=5{{q}^{2}}+6q+1$.
Here the square of the number is of the form $5m+4$.
From (5), $a=5q+4$.
Squaring both sides we get: ${{a}^{2}}={{\left( 5q+4 \right)}^{2}}\Rightarrow {{a}^{2}}=25{{q}^{2}}+16+40q$.
Splitting 16 into 15+1 and taking 5 common from $25{{q}^{2}}$,15,40q we get: ${{a}^{2}}=5\left( 5{{q}^{2}}+8q+3 \right)+1\Rightarrow {{a}^{2}}=5m+1$ where $m=5{{q}^{2}}+8q+3$.
Here the square of the number is of the form $5m+1$.
Now we can see from all numbers, none of the squares number is of the form $5m+2,5m+3$ where m is any integer.
Hence proved.

Note: Students should note that, we have used ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+ab$ for calculating the squared terms. Here, we just need to find the type of number squared numbers can have and then prove our result. Students can forget to take the value of r as 0.