If p mod 4 = 3, does choosing a=1 in the parameters of EdDSA really achieve best performance? I am implementing Ed448 in my crypto library. I found that if $a=-1$, I can use the formulas describing in this webpage, which saves some multiplications.

However, in the page 6 of RFC 8032, it says


  A non-zero square element $a$ of $GF(p)$.  The usual recommendation
  for best performance is $a = -1$ if $p \bmod 4 = 1$, and $a = 1$ if
  $p \bmod 4 = 3$.


Why does choosing $a=1$ if $p \bmod 4 = 3$ achieve the best performance?
There are probably better ways to do this, but at least this is quite elementary.

Let $p$ be an odd prime, and $\mathbb{F}_p^*=\langle\alpha\rangle$ be the cyclic subgroup of order $p-1$. Let $x\in\mathbb{F}_p^*$ such that $x^2=-1$. Take $t\in\{0,\ldots,p-2\}$ such that $x=\alpha^t$. Then $\alpha^{4t}=x^4=1$, so $4t\equiv 0\bmod p-1$. As $0\leq 4t\leq 4p-8$, there are 4 options:


$4t=0$. Then $t=0$, so $x=1$, a contradiction.
$4t=p-1$. Then $p=4t+1$, so $p\equiv 1\bmod 4$.
$4t=2p-2$. Then $t = (p-1)/2$, so $x^2 = 1$, a contradiction. 
$4t = 3p-3$. Then $3p\equiv 3\bmod 4$, so $p\equiv 1\bmod 4$.


In other words, $-1$ is not a square if $p\equiv 3\bmod 4$. This is an explicit assumption in your quote.