Prove that if n is even then $n^2$ is even and if n is odd then $n^2$ is odd?

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Solution 1

Hint:

$$n^2-n=n(n-1)$$ which is even being product of two consecutive integers

So, $n^2,n$ have the same parity.

More generally, we can prove $n, n^m( m\ge1)$ have the same parity.

Solution 2

Suppose $n$ is even. Let $p=n/2$.

$n^2=4p^2$ which must be even since $p$ is a whole number.

Suppose $n$ is odd and let p=$n+1$.

$p^2$ is therefore even by the above rule.

$n=(p-1)^2=p^2-2p+1$.

$p^2-2p$ must be even so $n$ is therefore odd.

Solution 3

  1. When n is even. Then n = 2k.

$n^2 = (2k)^2$

= $2(2k^2)$

= $2t$ where $t = 2k^2$

Which is same form as even numbers as you can see.

  1. When n is odd. Then n = 2k + 1

$n^2 = (2k + 1)^2$

= $4k^2 + 4k + 1$

= $2(2k^2 + 2k) + 1$

= $2t + 1$ where $t = 2k^2 + 2k$

Which is odd as you can see.

Solution 4

An integer is even iff it is of the form $2n$ and an integer is odd iff it is of the form $2n+1$. In both cases $n$ is an integer. Hence $$(2n)^2=4n^2=2(2n^2)$$ $$(2n+1)^2=4n^2+4n+1=2(2n^2+2n)+1$$ Since $n$ is an integer, both $2n^2$ and $2n^2+2n$ are integers.

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Updated on August 01, 2022

Comments

  • user407952
    user407952 over 1 year

    How can I prove that if $n$ is even then $n^2$ is even and if $n$ is odd then $n^2$ is odd?

  • Joffan
    Joffan almost 7 years
    Seems unnecessarily confusing.
  • Anonymous
    Anonymous almost 7 years
    I think it's misleading to call it just a hint :) Then again, this is a remarkably elegant proof.