Question

prove that difference of squares of two distinct odd natural numbers is always a multiple of 8 .
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Answer 1

Answer:

Concept used:

The difference of any two distinct odd natural numbers is even

Let 2m+1 and 2n+1 be two distinct odd natural numbrs with 2m+1 > 2n+1

Now,

$(2m+1)^2-(2n+1)^2$

$=[2m+1+2n+1][2m+1-(2n+1)]$

$=[2m+2n+2][2m+1-2n-1]$

$=2[m+n+1][2m-2n]$

$=2(m+n+1)*2(m-n)$

$=4(m+n+1)(m-n)$

clearly m-n is an even number

Then

$(2m+1)^2-(2n+1)^2=4(m+n+1)(2k)$

$(2m+1)^2-(2n+1)^2=8k(m+n+1)$

Hence $(2m+1)^2-(2n+1)^2$ is a multiple of 8

Answer 2

Step-by-step explanation:

The difference is divisible by 8. The two factors (n−m)+2m+1 and n−m differ by an odd number (2m+1), so they have opposite parity. Therefore, one of them is even, so their product is even so 4((n−m)+2m+1)(n−m) is divisible by 8.