using euclids division lemma show that the square of any positive integer is of the 5m, 5m+1 or 5m+4 for some integer m
Answers
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 .