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

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 ✅✅.

Answer 2

Let n be any positive integer.

By Euclid’s division lemma,

n = 5q + r, 0 ≤ r < 5

n = 5q, 5q + 1, 5q + 2, 5q + 3 or 5q + 4 (q is a whole number)

Now n^2 = (5q)^2 = 25q^2 = 5(5q^2) = 5m

n^2 = (5q + 1)^2 = 25q^2 + 10q + 1 = 5m + 1

n^2 = (5q + 2)^2 = 25q^2 + 20q + 4 = 5m + 4

Similarly,

n^2 = (5q + 3)^2 = 25q^2 + 30^q + 5 + 4

= 5m + 4

and n^2 = (5q + 4)^2 = 5m + 1

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