摘要: |
利用高斯二平方和定理求解一个特殊的丢番图方程〖SX(〗1〖〗x2〖SX)〗+〖SX(〗1〖〗y2〖SX)〗=〖SX(〗1〖〗z2〖SX)〗+〖SX(〗1〖〗w2〖SX)〗,将其转化为a2+b2=c2+d2.经讨论得知,a2+b2≡c2+d2≡1,2(mod 4),当(k1-k3)(k1+k3-1)≡(k4+k2)(k4-k2)时,a2+b2≡c2+d2≡1(mod 4);当(k1-k3)(k1+k3-1)≡(k4-k2)(k4+k2- |
关键词: 丢番图方程 高斯二平方和定理 整数解 |
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A Solution to the Diophantus Equation 〖SX(〗1〖〗x2〖SX)〗+〖SX(〗1〖〗y2〖SX)〗=〖SX(〗1〖〗z2〖SX)〗+〖SX(〗1〖〗w2 |
WANG Heng feng,CHEN Xing
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Abstract: |
In this article, the sum of two squares and Gauss theorem is used to solve a particular diophantus equation 〖SX(〗1〖〗x2〖SX)〗+〖SX(〗1〖〗y2〖SX)〗=〖SX(〗1〖〗z2〖SX)〗+〖SX(〗1〖〗w2〖SX)〗. 〖SX(〗1〖〗x2〖SX)〗+〖SX(〗1〖〗y2〖SX)〗=〖SX(〗1〖〗z2〖SX)〗+〖SX(〗1〖〗w2〖SX)〗 will be converted to a2+b2=c2+d2. After discussion,a2+b2≡c2+d2≡1,2(mod 4). We say that a2+b2≡c2+d2≡1(mod 4) if and only if (k1-k3)(k1+k3-1)≡(k4+k2)(k4-k2)≡0(mod 4); we say that a2+b2≡c2+d2≡2(mod 4) if and only if (k1-k3)(k1+k3-1)≡(k4-k2)(k4+k2-1)≡0,2(mod 4). |
Key words: diophantus equation the sum of two squares and Gauss theorem integer solution |