show that the squre of an odd positive integer is of the form 8m + 1 for some integer m.
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Answer:
hey there !
have a look at your answer....
Note :- I am taking q as some integer .
Let 'a' be the any positive integer.
Then, b = 8 .
Using Euclid's division lemma :-
0 ≤ r < b => 0 ≤ r < 8 .
•°• The possible values of r is 0, 1, 2, 3, 4, 5, 6, 7.
▶ Question said Square of odd positive integer , then r = 1, 3, 5, 7 .
→ Taking r = 1 .
a = bm + r .
= (8q + 1)² .
= 64m² + 16m + 1
= 8( 8m²+ 2m ) + 1 .
= 8q + 1 . [ Where q = 8m² + 2m ]
→ Taking r = 3 .
a = bq + r .
= (8q + 3)² .
= 64m² + 48m + 9 = 64m² + 48m + 8 + 1 .
= 8( 8m²+ 6m + 1 ) + 1 .
= 8q + 1 . [ Where q = 8m² + 6m + 1 ]
→ Taking r = 5 .
a = (8q + 5)² .
= 64m² + 80m + 25 = 64m² + 80m + 24 + 1 .
= 8( 8m²+ 10m + 3 ) + 1 .
= 8q + 1 . [ Where q = 8m² + 10m + 3 ]
→ Taking r = 7 .
a = ( 8q + 7 )² .
= 64m² + 112m + 49 = 64m² + 112m + 48 + 1 .
= 8( 8m²+ 14m + 6 ) + 1 .
= 8q + 1 . [ Where q = 8m² + 14m + 6 ] .
Hence, the square of any odd positive integer is of the form 8q + 1 .
✓✓ Proved ✓✓
You can see that at every odd values of r square of a is in the form of 8m +1
But at every even Values of r square of a the answer will not be in the form of 8m +1 .
Also we know, a = 8q +1 , 8q +3 , 8q + 5 , 9q +7 are not divisible by 2 means these all numbers are odd numbers
Hence , it is clear that square of an odd positive is in form of 8m +1
hope it helps you deaR!
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