Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q

Solution:

Consider a positive integer a = 4m + r ,

Using division algorithm we know that 0 ≤ r < 4 ,

When r = 0

a = 4m.

By squaring both sides,

a² = (4m)²

a² = 4 (4m²)

a² = 4 q , where q = 4m²

When r = 1

a = 4m + 1

By squaring both sides ,

a² = (4m + 1)²

a² = 16m² + 1 + 8m

a² = 4(4m² + 2m) + 1

a² = 4q + 1 , where q = 4m² + 2m

When r = 2

a = 4m + 2

By squaring both sides,

a² = (4m + 2)²

a² = 16m² + 4 + 16m

a² = 4(4m² + 4m + 1)

a² = 4q , Where q = 4m² + 4m + 1

When r = 3

a = 4m + 3

By squaring both sides ,

a² = (4m + 3)²

a² = 16m² + 9 + 24m

a² = 16m² + 24m + 8 + 1

a² = 4(4m² + 6m + 2) + 1

a² = 4q + 1 , where q = 4m² + 6m + 2

Therefore, the square of any positive integer is in the form of 4q or 4q + 1 , where q is any integer

✦ Try This: Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m

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

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NCERT Exemplar Class 10 Maths Exercise 1.3 Problem 1

Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q

Summary:

Square of any positive integer is in the form of 4q or 4q + 1 , where q is any integer

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☛ Related Questions:

• Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m
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• Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer . . . .