If 'n'is an odd positive integer, show that (n2-1) Is divisible by 8??
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10. Given that  prove that .
Proof: By the definition of divides a = be and c = df for some integers e and f. Thus ac=bedf = bdef=(bd)(ef). Thus  by the definition of divides.
14. Prove that if n is odd then n2 – 1 is divisible by 8.
Proof: An odd integer n is either a 4k+1 or a 4k+3.
(4k+1)2 – 1 = 16k2+ 8k + 1 - 1 =8(2k2 +k), which is a multiple of 8.
(4k+3)2 – 1 = 16k2+ 24k + 9 - 1 =8(2k2 +3k + 1), which is a multiple of 8.
21. Prove that if an integer is of the form 6k+5 it is necessarily of the form 3k-1, but not conversely.
Proof: 6k+5 = 6k + 5 + 1 – 1 = 6k + 6 – 1 = 3(2k+2) – 1, a number of the form 3k -1.
8 = 3(3) – 1 is of the form 3k-1, however, if 8 = 6k+5 then k = ½ which is not an integer. There is a number of the form 3k-1 that is not of the form 6k+5.
23. Prove that the square of any integer is of the form 3k or 3k+1 but not of the form 3k+2.
Proof: Every integer is of one of the three forms: 3k or 3k+1 or 3k+2.
Thus every square is of the form (3k)2 or (3k+1)2 or (3k+2)2.
(3k)2 = 3(3k2), a number of the form 3k.
(3k+1)2 = 9k2+6k+1=3(3k2+2k)+1, a number of the form 3k+1.
(3k+2)2 = 9k2 + 12k + 4 = 9k2 + 12k + 3 + 1 = 3(3k2 + 4k +1) + 1, a number of the form 3k+1.
Proof: By the definition of divides a = be and c = df for some integers e and f. Thus ac=bedf = bdef=(bd)(ef). Thus  by the definition of divides.
14. Prove that if n is odd then n2 – 1 is divisible by 8.
Proof: An odd integer n is either a 4k+1 or a 4k+3.
(4k+1)2 – 1 = 16k2+ 8k + 1 - 1 =8(2k2 +k), which is a multiple of 8.
(4k+3)2 – 1 = 16k2+ 24k + 9 - 1 =8(2k2 +3k + 1), which is a multiple of 8.
21. Prove that if an integer is of the form 6k+5 it is necessarily of the form 3k-1, but not conversely.
Proof: 6k+5 = 6k + 5 + 1 – 1 = 6k + 6 – 1 = 3(2k+2) – 1, a number of the form 3k -1.
8 = 3(3) – 1 is of the form 3k-1, however, if 8 = 6k+5 then k = ½ which is not an integer. There is a number of the form 3k-1 that is not of the form 6k+5.
23. Prove that the square of any integer is of the form 3k or 3k+1 but not of the form 3k+2.
Proof: Every integer is of one of the three forms: 3k or 3k+1 or 3k+2.
Thus every square is of the form (3k)2 or (3k+1)2 or (3k+2)2.
(3k)2 = 3(3k2), a number of the form 3k.
(3k+1)2 = 9k2+6k+1=3(3k2+2k)+1, a number of the form 3k+1.
(3k+2)2 = 9k2 + 12k + 4 = 9k2 + 12k + 3 + 1 = 3(3k2 + 4k +1) + 1, a number of the form 3k+1.
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