Can we argue that the square of every odd integer is of the form $8k+1$ using non inductive proof?
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Sure. If $n=2k+1$ then $n^2=4k^2+4k+1=4k(k+1)+1$ and either $k$ or $k+1$ is even, so $4k(k+1)$ is divisible by $8$.
carmichael561
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