Show that the square of any odd integer is of the form 4q + 1, for some integer q

Solution:

Consider a as any odd integer and b = 4.

Using Euclid's algorithm,

a = 4m + r for some integer m ≥ 0

r = 0, 1, 2, 3 as 0 ≤ r < 4.

We know that,

a = 4m or 4m + 1 or 4m + 2 or 4m + 3 So, a = 4m + 1 or 4m + 3.

a cannot be 4m or 4m + 2, as they are divisible by 2.

(4m + 1)² = 16m² + 8m + 1

= 4(4m² + 2m) + 1

= 4q + 1,

where q is some integer and q = 4m² + 2m.

(4m + 3)² = 16m² + 24m + 9

= 4(4m² + 6m + 2) + 1

= 4q + 1,

where q is some integer and q = 4m² + 6m + 2

Therefore, the square of any odd integer is of the form 4q + 1, for some integer q

✦ Try This: Show that the square of any odd integer is of the form 5p + 1, for some integer p

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 1

NCERT Exemplar Class 10 Maths Exercise 1.3 Problem 5

Show that the square of any odd integer is of the form 4q + 1, for some integer q

Summary:

The square of any odd integer is of the form 4q + 1, for some integer q. Hence proved

☛ Related Questions:

• If n is an odd integer, then show that n² – 1 is divisible by 8
• Prove that if x and y are both odd positive integers, then x² + y² is even but not divisible by 4
• Use Euclid’s division algorithm to find the HCF of 441, 567, 693