ISI2014-DCG-36

Question

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

• A. $I$ is never divisible by $2$
• B. $I$ is never divisible by $4$
• C. $I$ is never divisible by $6$
• D. None of the above

Answer 1

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$

Now,

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

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.