plz help me I am not getting the question I tried but too
Answers
It's so simple!!!
First write first a few positive odd integers.
1, 3, 5, 7, 9, 11, 13, 15, 17,...
Now take the squares of each.
1, 9, 25, 49, 81, 121, 169, 225, 289,...
Divide each squares by 6, and then write the remainders thus obtained.
1, 3, 1, 1, 3, 1, 1, 3, 1,...
The sequence thus formed is the repetition of the sequence 1, 3, 1.
Here we can find that the square of any odd positive integer leaves remainder either 1 or 3 on division by 6.
The squares of positive odd integers leaving remainder 1 on division by 6 can be written algebraically as 6q + 1. Also, such squares include in the AP of 'q'th term 6q + 1.
The squares of positive odd integers leaving remainder 3 on division by 6 can be written algebraically as 6q + 3. Also, such squares include in the AP of 'q'th term 6q + 3.
But in the repeating sequence formed by the remainders, i.e., 1, 3, 1, 1, 3, 1,... no 5 is seen. It means that none of the squares of odd positive integers leave remainder 5 on division by 6.
As there's none, there's no square of odd positive integer in the form 6q + 5.
Thus we get a conclusion that the square of any odd positive integer is in the form either 6q + 1 or 6q + 3 BUT NOR 6q + 5.