Prove that if n is even then $n^2$ is even and if n is odd then $n^2$ is odd?
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
- 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.
- 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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user407952
Updated on August 01, 2022Comments
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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?
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Joffan almost 7 yearsSeems unnecessarily confusing.
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Anonymous almost 7 yearsI think it's misleading to call it just a hint :) Then again, this is a remarkably elegant proof.