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Multiples of 8 plus 1
A variation is the difference of squares of any two odd numbers is always a multiple of 8. (2r-1)^2 = 4r^2 -2r +1 (2(r+n) -1)^2 = 4r^2 +8nr =4n^2 -4r -4n +1 The difference = 4n^2 +8nr - 4n 8nr has 8 as a factor. The remaining terms 4n^2-4n = 4n(n-1) Since n or n-1 has to be even, 4n(n-1) has to to be a multiple of 8 also hence as required! context |
Multiples of 8 plus 1
the last equals sign in the long line of working should be a + , sorry! context |