Prove if $n^2$ is even, then $n$ is even. I am just learning maths, and would like someone to verify my proof.
Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.
Is this valid?

We know that, 


*

*$n^2=n\times n $, We also know that 

*even $\times$ even = even

*odd  $\times$ odd  = odd

*odd  $\times$ even = even   


Observation $1$: As $n^2$ is even, we also get an even result in the 2nd and 4th case . 
Observation $2$: In the expression "$n \times n$" both operands are same i.e. '$n$', hence we get the result even in the $2$nd and $3$rd case  
Since the $2$nd case is common in both the operations we take the first case.Hence comparing even $\times$ even = even and $n^2=n\times n $,
Hence proved if $n^2$ is even, then $n$ is even.
