Prove that if m is an odd integer then there is an integer n such that...

Question

Question

Prove that if m is an odd integer then there is an integer n such that n= 4m+ 1 or n= 4m+ 3. Use a proof by cases.

Answer 1

Answer #1

If n is an odd integer. then n = 4m + 1 or n = 4m + 3 for some integer m, so that either

Solution :
n = 4m + 1 -----[1]
n = 4m + 3 -----[2]

Square the Equation [1] and [2]

n^2 = (4m + 1)^2
= 16m^2 + 8m + 1
= 8(2m + 1) + 1 ---[3]

n^2 = (4m + 3)^2
= 16m^2 + 24m + 1
   = 8(2m + 3) + 1 ---[4]


From Equation [3] amd [4] both cases having
n^2 = 8k + 1

Answer 2

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