Question

using euclid's lemma show that square of any positive ibteger cannot be of form 5m+2 or 5m+3 fpr some integer m​

Answer 1

Number divisible by 5 can be of the form:-

d = 5m + r,

where 0 ≤ r <5

If d = 5m, d² = 5.q,

where q is some integer and q = 5m²

If d = 5m + 1, d² = 5q + 1

If d = 5m + 2, d² = 5q + 4

If d = 5m + 3, d² = 5q + 4

If d = 5m + 4, d² = 5q + 1

Therefore, the square of any positive integer cannot be in the form of 5q + 2 or 5q + 3 for any integer "q".

Answer 2

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 .