Question

Prove that the square of any integer is of form $5q,5q+1,5q+4$ for some integer $q$.

Answer 1

Hint: In the above question, first compare the given term with Euclid Division Lemma, then take the value of reminder $r=0,1,2,3,4$ and $b=5$. At last take the square of each term and rearrange in the form of Euclid lemma.

Complete step-by-step solution:
For the given question, first take the Euclid Division Lemma and compare the given terms.
By Division Lemma, we have
$a=bm+r$, where$0\le r\ge 4$ …………….(i)
It means $r$is greater or equal to $0$ and less than or equal to $4$.
If we are putting $b=5$ then,
$a=5m+r$
Now we have to put values of $r$, the we get
If $r=0,$ then $a=5m$
If we take $r=1,\,\,then\,\,a=5m+1$
If we take $r=2,\,\,then\,a=5m+2$
If we take $r=3\,\,then\,\,a=5m+3$
And at last if we take $r=4\,\,then\,\,a=5m+4$
They all are integers, now we have to check whether their squares are integers or not.
Then, on taking square of each term and arranging in the form of Division lemma, we get
${{\left( 5m \right)}^{2}}=25{{m}^{2}}$
After rearranging,
$\Rightarrow$${{\left( 5m \right)}^{2}}=5(5{{m}^{2}})$$=$$5q$
Where $q$ is some integer.
Now on squaring of $5m+1$, we get
$\Rightarrow$${{(5m+1)}^{2}}=25{{m}^{2}}+10m+1$
On applying ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ in the above equation.
$\Rightarrow$${(5m+1)^2}=5(5m^2+2m)+1$
$\Rightarrow$${(5m+1)^2}=5q+1$
Where $q=5{{m}^{2}}+2m$ is some integer.
Now on squaring $5m+2$ with the help of same identity, we get
${{\left( 5m+2 \right)}^{2}}=25{{m}^{2}}+20m+4$
On rearranging in the form of Division Lemma,
$\Rightarrow$${{\left( 5m+2 \right)}^{2}}=5(5{{m}^{2}}+4m)+4$
$\Rightarrow$${{\left( 5m+2 \right)}^{2}}=5q+4$
Similarly $q$ is some integer.
Now on squaring $5m+3$ we get
$\Rightarrow$${{\left( 5m+3 \right)}^{2}}=25{{m}^{2}}+30m+9$
We can write $9$ as $5+4$ so that
$\Rightarrow$${{\left( 5m+3 \right)}^{2}}=25{{m}^{2}}+30m+5+4$
On rearranging in the form of Division lemma, we get
$\Rightarrow$${{\left( 5m+3 \right)}^{2}}=5(5{{m}^{2}}+6m+1)+4$
$\Rightarrow$${{\left( 5m+3 \right)}^{2}}=5q+4$
Where q is some integer.
Now on squaring $5m+4$, we get
$\Rightarrow$${{\left( 5m+4 \right)}^{2}}=25{{m}^{2}}+40m+16$
We can write $16$ as $15+1$, so that the above equation
$\Rightarrow$${{\left( 5m+4 \right)}^{2}}=25{{m}^{2}}+40m+15+1$
On rearranging in the form of Division Lemma , we get
$\Rightarrow$${{\left( 5m+4 \right)}^{2}}=5(5{{m}^{2}}+8m+3)+1$
And we write it as
$\Rightarrow$${{\left( 5m+4 \right)}^{2}}=5q+1$
Similarly $q$ is some integer.

Hence, the square of any positive integer is of the form $5q\,\,or\,\,5q+1\,\,or\,\,5q+4$ for some integer $q$.

Note: Euclid division lemma- If we have two positive integers a and b , then there would be a whole number q and r that satisfy the equation $a=bq+r $.
For this type of question we have to rearrange the given lemma.
In such types of problems, students are unable to distinguish between Euclid Division Lemma and Euclid's Algorithm and they make mistakes. The Euclid Division Algorithm is related to LCM and HCF.
Euclid Division Algorithm- This algorithm is the process of applying Euclid’s division lemma in succession several times to obtain the HCF of any two numbers.