Question

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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.

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Answer 1

▶ 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 ✅✅.



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Answer 2

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.


Let a be any positive integer and b= 5.

According to Euclid's Division lemma,

=) a = bq + r, where r = 0,1,2,3,4.

=) a = 5q, 5q+1, 5q+2, 5q+3, 5q+4.

If a = 5q,

=) a² = (5q)²

= 25q²

= 5(m) where m = 5q².

If a = 5q+1,

=) a² = (5q+1)²

= 25q² + 10q + 1

= 5(5q² + 2q) +1

= 5m + 1, where m = 5q² + 2q.

If a = 5q+2,

=) a² = (5q+2)²

= 25q² + 20q + 4

= 5(5q² + 4q) + 4

= 5m + 4, where m = 5q² + 4q.

If a = 5q+ 3,

=) a² = (5q+3)²

= 25q² + 30q + 9

= 25q² + 30q + 5 + 4

= 5(5q² + 6q +1)+ 4

= 5m + 4, where m = (5q² + 6q +1).

If a = 5q + 4,

=) a² = (5q+4)²

= 25q² + 40q + 16

= 25q² + 40q + 15 + 1

= 5(5q² + 8q +3) + 1

= 5m +1, where m = (5q² + 8q +3).

Hence Square of any positive integer cannot be of form 5m+2 or 5m+3 for some integer m. Proved.