Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m

Solution:

Consider a positive integer = a

Using the Euclid’s division algorithm,

a = 6q + r, where 0 ≤ r < 6

a² = (6q + r)² = 36q² + r² + 12qr  [∵(a + b)² = a² + 2ab + b²]

a² = 6(6q² + 2qr) + r² --- (i), where,0 ≤ r < 6.

When r = 0,

a² = 6 (6q²) = 6m,

where m = 6q² is an integer.

When r = 1,

a² = 6 (6q² + 2q) + 1 = 6m + 1,

where m = (6q² + 2q) is an integer.

When r = 2,

a² = 6(6q² + 4q) + 4 = 6m + 4,

where m = (6q² + 4q) is an integer.

When r = 3,

a² = 6(6q² + 6q) + 9 = 6(6q² + 6q) + 6 + 3

a² = 6(6q² + 6q + 1) + 3 = 6m + 3,

where m = (6q + 6q + 1) is integer.

When r = 4,

a² = 6(6q² + 8q) + 16

= 6(6q² + 8q) + 12 + 4

a² = 6(6q² + 8q + 2) + 4 = 6m + 4,

where m = (6q² + 8q + 2) is integer.

When r = 5,

a² = 6 (6q² + 10q) + 25

= 6(6q² + 10q) + 24 + 1

a² = 6(6q² + 10q + 4) + 1 = 6m + 1,

where m = (6q² + 10q + 4) is integer

Therefore, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m

✦ Try This: Show that the square of any positive integer cannot be of the form 4m + 2 or 4m + 5 for any 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 4

Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m

Summary:

The square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m. Hence proved

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☛ Related Questions:

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