Respuestas
¿Demasiados anuncios?Como usted ha dicho, tenemos que usar un truco completamente diferente de 1 y 2. Si pensamos en la ecuación de $\mathbb{C}$, entonces sabemos que that$$x^2 + y^2 = (x + \sqrt{-1}y)(x-\sqrt{-1}y).$$Next, as we know that $\sqrt{-1}$ is in $\mathbb{Z}_5$ (use Hensel's lemma with $2^2 = - 5\text{)}$), then we know that in $\mathbb{Z}_5$ this factorization also exists. Hence, given any nonzero $x$ in $\mathbb{Z}_5$, I can give you two different $y$ in $\mathbb{Z}_5$ such that $x^2 {(mod} 1\text + y ^ 2 = 0 $, while if $x = 0 $, then we have $y = 0$.