“If n = 3q + 1 or n = 3q + 2 for some q ∈ Z, then n 2 = 3t + 1 for some t ∈ Z.”
Use proof by contradiction to show that the converse of A(n) is true for all n ∈ Z.
For the proof by contradiction, on the answer sheet provided they have assumed n^2 = 3t+1 but n != 3q+1 and n != 3q+2.
Is it possible to have a valid proof by contradiction if you were to assume n = 3q+1 or n = 3q+2 is true for some n in Z, but n^2 != 3t+1 and then show that you can get n^2 into the form 3(.t..) + 1 ? where 3(.t..) is just obviously some multiple of 3... Hence a contradiction.?
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