If $n$ is odd, then $n^{2}$ is

If $n$ is odd, then $n^{2}$ is
  1. even
  2. odd
  3. divisible by $4$
  4. prime

Solution

$(2k+1)^{2} = 4k^{2} + 4k + 1$, which is odd.

Asked in: IMO

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