\(\left(4n+3\right)^2-25=\left(4n+3-5\right)\left(4n+3+5\right)\)
\(=\left(4n-2\right)\left(4n+8\right)=2\left(2n-1\right).4\left(n+2\right)\)
\(=8\left(2n-1\right)\left(n+2\right)⋮8\forall n\in Z\)
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