prove that the result obtained by subtraction of a square of an integer is a multiple of 8
Answers
Given : Result obtained by subtraction of 1 from a square of an odd integer is a multiple of 8 (Correction in Question )
To find : Prove that
Solution:
odd integer = 2n+ 1
Square = (2n + 1)²
Square - 1
= (2n + 1)² - 1
= (2n + 1)² - 1²
Using a² - b² = (a + b)(a - b)
= (2n + 1 + 1)(2n +1 - 1)
= (2n + 2)(2n)
= 2(n + 1)(2(n))
= 4 n(n + 1)
Any number n can be of form
2k or 2k + 1
n = 2k
4 n(n + 1)
= 4 (2k)(2k + 1)
= 8k(2k+1)
Divisible by 8
n = 2k + 1
4 n(n + 1)
= 4 (2k+1)(2k + 1 +1)
= 4 (2k+1)(2k +2)
= 4 (2k+1)2(k +1)
= 8(2k+1)(k + 1)
Divisible by 8
QED
Hence proved
Result obtained by subtraction of 1 from a square of an odd integer is a multiple of 8
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