Prove that the difference of squares of 2 odd natural nos. is a multiple of 8

(2m+1)^2-(2n+1)^2
(4m^2+4m+1)-(4n^2+4n+1)
4m^2+4m-4n^2-4n
4[m^2+m-n^2-n]
4[(m+1)m-(n+1)n]

either m or (m+1) is even same with n difference of even is even even times 4 is multiple of 8

Or you could note that the difference between the squares of consecutive odd numbers is:
(2n + 1)² - (2n - 1)² = (4n² + 4n + 1) - (4n² - 4n + 1) = 8n is divisible by 8,
from which the result immediately follows.

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