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Consider any integer $I=m^2+n^2$, where $m$ and $n$ are odd integers. Then

  1. $I$ is never divisible by $2$
  2. $I$ is never divisible by $4$
  3. $I$ is never divisible by $6$
  4. None of the above
in Numerical Ability by Veteran (434k points)
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1 Answer

+1 vote
Answer $B$

Since, $m$ and $n$ are odd numbers. So, they can be expressed as:

$m = 2x+1, x\in Z$, and $ n = 2y + 1, y\in Z$


$m^2 + n^2 = (2x+1)^2 + (2y+1)^2 = 4x^2+1+4x + 4y^2+1+4y = 4 \underbrace {(x^2+x+y^2+y)}_k + 2 = 4k+2$

So, $m^2+n^2$ can be expressed as $m^2 + n^2 = 4k+2$, where $k \in Z$

for $k =\pm0, \pm1, \pm2, \pm3\cdots$

So, values obtained are: $0, \pm6, \pm14, \pm18. \cdots$

Hence, It will never be divisible by $4$.

Therefore, option $B$ is the correct answer.
by Boss (19.3k points)
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