Question

use euclid division Lemma to show that the square of any positive integer cannot be in the form of 5 M + 2 upon 3 or 5 M + 3 for some integer m

Answer 1

Sol : Let x be any positive Integer Then x = 5q or x = 5q+1 or x = 5q+4  for integer x. If x = 5q, x2 = (5q)2 = 25q2 = 5(5q2) = 5n (where n = 5q2 ) If x = 5q+1, x2 = (5q+1)2= 25q2+10q+1 = 5(5q2+2q)+1 = 5n+1 (where n = 5q2+2q ) If x = 5q+4, x2 = (5q+4)2 = 25q2+40q+16 = 5(5q2 + 8q + 3)+ 1 = 5n+1 (where n = 5q2+8q+3 )∴in each of three cases x2 is either of the form 5q or 5q+1 or 5q+4 and for integer q.

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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.

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▶ 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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