1.若s,t∈S,
則s = m^2 + n^2,t = u^2 + v^2,其中m,n,u,v∈Z.
那么st = (m^2 + n^2)(u^2 + v^2) = (mu + nv)^2 + (mv - nu)^2,
其中mu + nv∈Z.,mv - nu∈Z.
所以st∈S
2.若s,t∈S,t ≠0,
仍設(shè)s = m^2 + n^2,t = u^2 + v^2,其中m,n,u,v∈Z.
因為t ≠0,故u,v不同時為零.
則s/t = st/t^2 = (m^2 + n^2)(u^2 + v^2)/(u^2 + v^2)^2
= ((mu + nv)^2 + (mv - nu)^2)/(u^2 + v^2)^2
= {(mu + nv)/(u^2 + v^2)}^2 + {(mv - nu)/(u^2 + v^2)}^2
設(shè)p = (mu + nv)/(u^2 + v^2),q = (mv - nu)/(u^2 + v^2),
則p,q為有理數(shù),且s/t=p²+q².