Find the mistakes in the “proofs” shown.
Exercise
Theorem: For all integers k, if k > 0 then k2 + 2k + 1 is composite.
“Proof: Suppose k is any integer such that k > 0. If k2 + 2k + 1 is composite, then k2 + 2k + 1 = rs for some integers r and s such that
1<r <(k2 + 2k + 1)
and 1<s <(k2 + 2k + 1).
Since k2 + 2k + 1 = rs
and both r and s are strictly between 1 and k2 + 2k + 1, then k2 + 2k + 1 is not prime. Hence k2 + 2k + 1 is composite as was to be shown.”
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