Question

show that the square of an odd integer is of the form 4q + 1,or 4q+3

Answer 1

Using Euclid division lemma we have,
a=bq +r ,where r is less than b
taking b = 4 we have
a = 4m
a = 4m + 1
a = 4m + 2 and
a = 4m + 3
since odd integers are of form 2n + 1 therefore numbers
a = 4m + 1 and a = 4m + 3 are odd integers
on squaring both side in equation a = 4m + 1 we have,
a^2 = 16m^2 + 8m + 1
thus we have,
a^2 = 4(4m^2 + 2m) + 1 on taking 4 common,
thus in substituting 4m^2 + 2m as q we have,
a^2 = 4q + 1 = square of odd number,
similarly on doing squaring in second equation we will get
a^2 = 4q + 3 = square of odd number hence in general, Any square of odd number is of this form

Answer 2

Step-by-step explanation:

Let a be the positive integer.

And, b = 4 .

Then by Euclid's division lemma,

We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .

°•° Then, possible values of r is 0, 1, 2, 3 .

Taking r = 0 .

a = 4q .

Taking r = 1 .

a = 4q + 1 .

Taking r = 2

a = 4q + 2 .

Taking r = 3 .

a = 4q + 3 .

But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .

•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .

Hence , it is solved .