Question

Use Euclid's division lemma to show that the square of any positive integer is of the form 3p,3p+1.​

Answer 1

Answer:

let us take, 'x'= 3q , 3q+1, 3q+2

when, x=3q

        x2 =  (3q) 2

         x2 = 9q2 

        x2  = 3(3q2)

we see that 3q2= m

so we have done the first equation 3m

when , x=3q+1

           x2= (3q+1)2

                                 [since, (a+b)2 = a2+2ab+b2]

           x2= 9q+6q+1

           x2= 3(3q+2q)+1

in this we see that 3q+2q= m

    therefore, this satisfy the equation m+1

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Anshita , Student

Member since Feb 04 2015

let us take, 'x'= 3q , 3q+1, 3q+2

when, x=3q

        x2 =  (3q) 2

         x2 = 9q2 

        x2  = 3(3q2)

we see that 3q2= m

so we have done the first equation 3m

when , x=3q+1

           x2= (3q+1)2

                                 [since, (a+b)2 = a2+2ab+b2]

           x2= 9q+6q+1

           x2= 3(3q+2q)+1

in this we see that 3q+2q= m

    therefore, this satisfy the equation m+1

⚡Hope it will help you.⚡

Answer 2

Step-by-step explanation:

Question : -

→ Use Euclid's Division lemma to show that the Square of any positive integer cannot be of form 5m + 2 or 5m + 3 for some integer m.

$\huge \pink{ \mid{ \underline{ \overline{ \tt Answer: -}} \mid}}$

▶ Step-by-step explanation : -

Let ‘a’ be the any positive integer .

And, b = 5 .

→ Using Euclid's division lemma :-

==> a = bq + r ; 0 ≤ r < b .

==> 0 ≤ r < 5 .

•°• Possible values of r = 0, 1, 2, 3, 4 .

→ Taking r = 0 .

Then, a = bq + r .

==> a = 5q + 0 .

==> a = ( 5q )² .

==> a = 5( 5q² ) .

•°• a = 5m . [ Where m = 5q² ] .

→ Taking r = 1 .

==> a = 5q + 1 .

==> a = ( 5q + 1 )² .

==> a = 25q² + 10q + 1 .

==> a = 5( 5q² + 2q ) + 1 .

•°• a = 5m + 1 . [ Where m = 5q² + 2q ] .

→ Taking r = 2 .

==> a = 5q + 2 .

==> a = ( 5q + 2 )² .

==> a = 25q² + 20q + 4 .

==> a = 5( 5q² + 4q ) + 4 .

•°• a = 5m + 4 . [ Where m = 5q² + 4q ] .

→ Taking r = 3 .

==> a = 5q + 3 .

==> a = ( 5q + 3 )² .

==> a = 25q² + 30q + 9 .

==> a = 25q² + 30q + 5 + 4 .

==> a = 5( 5q² + 6q + 1 ) + 4 .

•°• a = 5m + 4 . [ Where m = 5q² + 6q + 1 ] .

→ Taking r = 4 .

==> a = 5q + 4 .

==> a = ( 5q + 4 )² .

==> a = 25q² + 40q + 16 .

==> a = 25q² + 40q + 15 + 1 .

==> a = 5( 5q² + 8q + 3 ) + 1 .

•°• a = 5m + 1 . [ Where m = 5q² + 8q + 3 ] .

→ Therefore, square of any positive integer in cannot be of the form 5m + 2 or 5m + 3 .

✔✔ Hence, it is proved ✅✅.

$\huge \orange{ \boxed{ \boxed{ \mathscr{THANKS}}}}$