Question

Use Euclid division lemma to show that square of any positive integer cannot be of the from 5m+2 or 5m+3 for some integer m. (plz provide the whole workout)

Answer 1

Let the number be a which when divided by 5 gives q as quotient and r as remainder.

Then by Euclid's division lemma
that is
=> a = bq+r

where, b= 5
and r =0,1,2,3,4

a = 5q

a = 5q+1

a = 5q+2

a = 5q+3

a = 5q+4

Now taking all as a case:-

⭐Case -1

a = 5q

a² = (5q)²

a² = 25q²

a² = 5m { where m = 5q²}


⭐Case -2

a = 5q+1

a² = (5q+1)²

a² = 25q²+1+10q

[using identity (a+b)² = a² + b² + 2ab]

a² = 5m(where m= 5q²+2q)+1

a² = 5m+1

⭐Case -3

a= 5q+2

a² = (5q+2)²

a² = 25q² +4+20q

a²= 5m ( where m = 5q²+4q)+4

a² = 5m+4

⭐Case -4

a = 5q+3

a² = (5q+3)²

a² = 25q²+9+30q

a²=5m ( where m = 5q²+6q)+9

a²=5m+9

⭐Case -5

a =5q+4

a²=(5q+4)²

a²= 25q²+16+40q

a² = 5m ( where m= 5q²+8q)+16

a²= 5m+16



Since, by solving all we didn't get
a²=5m+2 Or a²=5m+3

So,proved.

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 ✅✅.

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

_'fff