已知S是两个整数平方和的集合,即S={X|X=M²+N²,M∈Z,n∈Z}.求证:1.若s,t∈S,则st∈S2.若s,t∈S,t≠0,则s/t=p²+q²,其中p,q为有理数
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![已知S是两个整数平方和的集合,即S={X|X=M²+N²,M∈Z,n∈Z}.求证:1.若s,t∈S,则st∈S2.若s,t∈S,t≠0,则s/t=p²+q²,其中p,q为有理数](/uploads/image/z/5303153-65-3.jpg?t=%E5%B7%B2%E7%9F%A5S%E6%98%AF%E4%B8%A4%E4%B8%AA%E6%95%B4%E6%95%B0%E5%B9%B3%E6%96%B9%E5%92%8C%E7%9A%84%E9%9B%86%E5%90%88%2C%E5%8D%B3S%3D%EF%BD%9BX%7CX%3DM%26sup2%3B%2BN%26sup2%3B%2CM%E2%88%88Z%2Cn%E2%88%88Z%EF%BD%9D.%E6%B1%82%E8%AF%81%EF%BC%9A1.%E8%8B%A5s%2Ct%E2%88%88S%2C%E5%88%99st%E2%88%88S2.%E8%8B%A5s%2Ct%E2%88%88S%2Ct%E2%89%A00%2C%E5%88%99s%2Ft%3Dp%26sup2%3B%2Bq%26sup2%3B%2C%E5%85%B6%E4%B8%ADp%2Cq%E4%B8%BA%E6%9C%89%E7%90%86%E6%95%B0)
已知S是两个整数平方和的集合,即S={X|X=M²+N²,M∈Z,n∈Z}.求证:1.若s,t∈S,则st∈S2.若s,t∈S,t≠0,则s/t=p²+q²,其中p,q为有理数
已知S是两个整数平方和的集合,即S={X|X=M²+N²,M∈Z,n∈Z}.求证:
1.若s,t∈S,则st∈S
2.若s,t∈S,t≠0,则s/t=p²+q²,其中p,q为有理数
已知S是两个整数平方和的集合,即S={X|X=M²+N²,M∈Z,n∈Z}.求证:1.若s,t∈S,则st∈S2.若s,t∈S,t≠0,则s/t=p²+q²,其中p,q为有理数
1.若s,t∈S,
则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,
仍设s = m^2 + n^2,t = u^2 + v^2,其中m,n,u,v∈Z.
因为t ≠0,故u,v不同时为零.
则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
设p = (mu + nv)/(u^2 + v^2),q = (mv - nu)/(u^2 + v^2),
则p,q为有理数,且s/t=p²+q².
证明:太难了...
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