Prove that for any integer n, the number n2 is even if n is even, and n2 is odd if n is odd. Proof: Consider two cases:
Case 1 • n is even • Then n=2k for some integer k • Then n2= (2k)2 = 4k2 = 2(2k2), which is even
Case 2 • n is odd • Then n = 2k+1 for some integer k • Then n2 = (2k+1)2 = 4k2 + 4k +1 = 2(2k2 + 2k) +1, which is odd Since both cases have been proven, the statement holds for all integers n. Prove that if an
integer is a
perfect cube, then it must be either a multiple of 9, 1 more than a multiple of 9, or 1 less than a multiple of 9.
Proof: Each perfect cube is the cube of some integer
n, where
n is either a multiple of 3, 1 more than a multiple of 3, or 1 less than a multiple of 3. So these three cases are exhaustive: • Case 1: If
n = 3
p, then
n3 = 27
p3, which is a multiple of 9. • Case 2: If
n = 3
p + 1, then
n3 = 27
p3 + 27
p2 + 9
p + 1, which is 1 more than a multiple of 9. For instance, if
n = 4 then
n3 = 64 = 9×7 + 1. • Case 3: If
n = 3
p − 1, then
n3 = 27
p3 − 27
p2 + 9
p − 1, which is 1 less than a multiple of 9. For instance, if
n = 5 then
n3 = 125 = 9×14 − 1.
Q.E.D. ==Elegance==