Prove that for every integer $x$, if $x$ is odd then there exists an integer $y$ such that $x^2=4y+1$. 
Prove that for every integer $x$, if $x$ is odd then there exists an integer $y$ such that $x^2=4y+1$.

Let $x$ be an odd integer. Then, there exists an integer m such that $x=2m+1$.  But $x^2=4m^2+4m+1=4(m^2+m)+1$.
Case 1: $m$ is odd. Then there exists a $k$ such that $m=2k+1$. Since $m^2=4k^2+4k+1$, $m^2$ is odd. Hence, $m^2+m$ is even and $m^2+m=2y$ for some $y$. Thus, $4(m^2+m)+1=4(y)+1$ as required.

As stated in the comment, you can take $y=m^2+m$ and you are done.
Alternatively, we just have to illustrate that $x^2-1$ is a multiple of $4$. Notice that $$x^2-1=(x-1)(x+1)$$
If $x$ is odd, both $x-1$ and $x+1$ are even, hence the product is a multiple of $4$.
