Question

Prove that, if x and y are odd positive integers, then ${{x}^{2}}+{{y}^{2}}$ is even but not divisible by 4.

Answer 1

Hint: Here, we will first consider x and y as $2m+1$ and $2n+1$. We will use the concept that an odd integer is not divisible by 2 and an even integer is divisible by 2. Then, we will try to prove the required result using the condition given in the problem.


Complete step-by-step solution -
Since, we know that when any integer is divided by 2, the remainder obtained is 0 or 1. So, any integer which is not divisible by 2 is of the form $2m +1$, where ‘m’ is an integer.
So, let us consider that $x = 2m +1$ and $y = 2n+1$.
Now, by Euclid’s division lemma, we know that if an integer ‘a’ is divided by another integer ‘b’, then we can write the integer ‘a’ as $a = bq+r$, where ‘q’ is the quotient and r is the remainder. The value of ‘r’ always lies between 0 and b, it can be equal to 0 but not equal to b.
Since, we have to show that ${{x}^{2}}+{{y}^{2}}$ is an even integer.
Since, $x = 2m+1$ and $y = 2n+1$, ${{x}^{2}}+{{y}^{2}}$ is given as:
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}={{\left( 2m+1 \right)}^{2}}+{{\left( 2n+1 \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=\left( 4{{m}^{2}}+1+4m \right)+\left( 4{{n}^{2}}+1+4n \right) \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=4\left( {{m}^{2}}+{{n}^{2}}+m+n \right)+2 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=2\left[ 2\left( {{m}^{2}}+{{n}^{2}}+m+n \right)+1 \right] \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=2k \\
\end{align}$
Here, $k=\left[ 2\left( {{m}^{2}}+{{n}^{2}}+m+n \right)+1 \right]$
As ${{x}^{2}}+{{y}^{2}}$ is of the form 2k, this means that it is divisible by 2 and hence it is even.
Now, any integer when divided by 4 will be written as $4q+r$ ( Using Euclid’s division lemma).
So, we have:
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}=4\left( {{m}^{2}}+{{n}^{2}}+m+n \right)+2 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=4k'+2 \\
\end{align}$
Here, $k'={{m}^{2}}+{{n}^{2}}+m+n$
Therefore, the value of r is 2. This means that we get a remainder 2, when ${{x}^{2}}+{{y}^{2}}$ is divided by 4. So, ${{x}^{2}}+{{y}^{2}}$ is not divisible by 4.
Hence, it is proved that if, x and y are odd positive integers, then ${{x}^{2}}+{{y}^{2}}$ is even but not divisible by 4.

Note: Students must remember the by Euclid’s division lemma, any number when divided by 2 can be expressed in the form of $2m + r$, where $r =0$ or $1$. It should also be kept in mind that an odd integer is always expressed in the form $2m+1$, where ‘m’ is an integer.