Prove that if $n$ is an integer, then $1 − n$ is even if and only if $n^2 + 1$ is even.

Question

I am practicing exam questions and have come across the following:
Prove that if n is an integer, then 1 − n is even if and only if $n^2$ + 1 is even.
The first thing that came to mind is a contrapositive proof, but how would I prove this?

Answer 1

For the forward direction:

$1-n$ being even would imply that $n$ is odd. Thus, $n^2$ is odd, so $n^2+1$ is even.

Backwards direction:

If $n^2+1$ is even, then $n^2$ is odd. Thus, $n$ is odd. Hence, $1-n$ is even.

For further clarity:

$n$ being odd means that $n = 2k+1$ for some integer $k$.

$n$ being even means that $n = 2k$ for some integer $k$.

You can proceed with a more formal argument using the above definitions.

Answer 2

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

If $n-1$ is even, $n-1+2 = n+1$ is even, their product is even and an even number plus 2 is even.

Going the other way, if $n^2 + 1$ is even, then so is $n^2 - 1.$ one of the factors $(n-1)(n+1)$ must be even. But if one is even the other is even.

Answer 3

You can do it straight forwardly in cases.

Either $n$ is even or odd.

Case 1: $n = 2m$ is even so $1-n =1- 2m = 2(-m) + 1$ is odd and $n^2 + 1 = 4m^2 + 1 = 2(2m^2) + 1$ is odd.

Neither $1-n$ nor $n^2 + 1$ is even.

Case 2: $n = 2m + 1$ is odd. So $1-n= 1-(2m + 1) = -2m = 2(-m)$ is even, and $n^2 + 1 = (2m+1)^2 + 1 = 4m^2 + 4m + 1 + 1 = 2(2m^2 + 2m +1)$ is even.

So both $1-n$ and $n^2 + 1$ are even.

So $1-n$ is even precisely and only when $n^2 + 1$ is even.

Answer 4

Hint: $\,\ 2\mid n(n\!+\!1) =\, n^2\, +\, 1\,\ -\,\ (1\,-\,n)\phantom{I_{I_{I_{I_1}}}}\,$
which implies that $\,2\mid n^2\!+\!1\!\iff\! 2\mid 1\!-\!n\ \ \ $ [here $\ a\mid b\ $ means $\,a\,$ divides $\,b\,$]

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Generally: $ $ if $\,d\mid y-x\,$ then $\ d\mid y\iff d\mid x$

i.e. $\bmod d\!:\, $ if $\ y\equiv x\,$ then $\,y\equiv 0\iff x\equiv 0$

i.e. on a line $\,y\equiv x$ we have $\,y\equiv 0\iff x\equiv 0$

Answer 5

$1 - n$ is even iff $n - 1$ is even iff $n$ is odd iff $n^2$ is odd iff $n^2 + 1$ is even.