Prove : Every even number can be written as the sum of an odd number and a perfect square
I'm defining an even number as $2n$, $n$ being any integer
For odd numbers, $2m + 1$, $m$ being any integer
For perfect squares, $k^2$, $k$ being any integer
Then, $2n = 2m + 1 + k^2$
not quite sure where to go from here...
some things to note are that these numbers can be positive or negative and
I've tried to disprove it with some simple numbers to no avail
even number = perfect square + odd number
$0 = 1 + (-1)$
$2 = 9 + (-7)$
$4 = 9 + (-5)$
$6 = 9 + (-3)$
$8 = 9 + (-1)$
$10 = 9 + 1$
$12 = 9 + 3$
and so on so it seems like it holds up
Any number can be written as the sum of a square and an odd number, start with $1=0+1$. If the number is a square greater than 1, then it can be written as $n^2+(2n+1)$, where $n\geq1$. Otherwise the number must be between two consecutive squares, say $n^2$ and $(n+1)^2$. All such numbers are of the form $n^2+k$, for some $k$. If $k$ is odd we're done, otherwise if $k$ is even, note that $k+2n-1$ is odd, and $n^2+k = (n-1)^2+(k+2n-1)$.