Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Show that the square of any positive integer cannot be of the form $6m+ 2$ or $6m + 5$ for any integer $m$.
Given:
"Square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any positive integer $m$".
To do:
We have to prove the given statement.
Solution:
According to Euclid's division lemma;
If $a$ and $b$ are two positive integers;
- $a\ =\ bq\ +\ r$, where $0\ \underline{< }\ r\ <\ b$.
If $b\ =\ 6$, then;
- $a\ =\ 6q\ +\ r$, where $0\ \underline{< }\ r\ <\ 6$.
- So, $r\ =\ 0,\ 1,\ 2,\ 3,\ 4,\ 5$
When, $r\ =\ 0$:
$a\ =\ 6q$
Squaring on both sides, we get:
$a^2\ =\ (6q)^2$
$a^2\ =\ 36q^2$
$a^2\ =\ 6(6q^2)$
$a^2\ =\ 6m$, where $m\ =\ 6q^2$
When, $r\ =\ 1$:
$a\ =\ 6q\ +\ 1$
Squaring on both sides, we get:
$a^2\ =\ (6q\ +\ 1)^2$
$a^2\ =\ 36q^2\ +\ 12q\ + 1$
$a^2\ =\ 6(6q^2\ +\ 2q)\ +\ 1$
$a^2\ =\ 6m\ +\ 1$, where $m\ =\ 6q^2\ +\ 2q$
When, $r\ =\ 2$:
$a\ =\ 6q\ +\ 2$
Squaring on both sides, we get:
$a^2\ =\ (6q\ +\ 2)^2$
$a^2\ =\ 36q^2\ +\ 24q\ + 4$
$a^2\ =\ 6(6q^2\ +\ 4q)\ +\ 4$
$a^2\ =\ 6m\ +\ 4$, where $m\ =\ 6q^2\ +\ 4q$
When, $r\ =\ 3$:
$a\ =\ 6q\ +\ 3$
Squaring on both sides, we get:
$a^2\ =\ (6q\ +\ 3)^2$
$a^2\ =\ 36q^2\ +\ 36q\ + 9$
$a^2\ =\ 36q^2\ +\ 36q\ + 6\ +\ 3$
$a^2\ =\ 6(6q^2\ +\ 6q\ +\ 1)\ +\ 3$
$a^2\ =\ 6m\ +\ 3$, where $m\ =\ 6q^2\ +\ 6q\ +\ 1$
When, $r\ =\ 4$:
$a\ =\ 6q\ +\ 4$
Squaring on both sides, we get:
$a^2\ =\ (6q\ +\ 4)^2$
$a^2\ =\ 36q^2\ +\ 48q\ + 16$
$a^2\ =\ 36q^2\ +\ 48q\ + 12\ +\ 4$
$a^2\ =\ 6(6q^2\ +\ 8q\ +\ 2)\ +\ 4$
$a^2\ =\ 6m\ +\ 4$, where $m\ =\ 6q^2\ +\ 8q\ +\ 2$
When, $r\ =\ 5$:
$a\ =\ 6q\ +\ 5$
Squaring on both sides, we get:
$a^2\ =\ (6q\ +\ 5)^2$
$a^2\ =\ 36q^2\ +\ 60q\ + 25$
$a^2\ =\ 36q^2\ +\ 60q\ + 24\ +\ 1$
$a^2\ =\ 6(6q^2\ +\ 10q\ +\ 4)\ +\ 1$
$a^2\ =\ 6m\ +\ 1$, where $m\ =\ 6q^2\ +\ 10q\ +\ 4$
Hence, the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any positive integer $m$.
30 Views
