Question

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

Answer 1

$\huge \bf \pink{ \mathcal{H}ey \: there !! }$




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

Let ‘a’ be the any positive integer .

And, b = 5 .

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 ] .



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