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素数公式,素数判断,求证:c=p,t=v

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发表于 2024-11-20 11:42 | 显示全部楼层 |阅读模式
已知:\(a^2+abc-abt=ct\),\(a=ct\),\(c\ne\sqrt{k+1}\),\(c\ne m^2-2\)
\(t\ne\sqrt{u+1}\),\(t\ne y^2-2\),整数\(a\ne0\),\(b\ne0\),\(k>0\)
\(m>0\),\(u>0\),\(y>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2+abc-abt=ct\),\(a=ct\),\(c\ne\sqrt{k+1}\),\(t\ne\sqrt{u+1}\)
整数\(a\ne0\),\(b\ne0\),\(k>0\),\(u>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2+abc-abt=ct\),\(a=ct\),\(c\ne m^2-2\),\(t\ne y^2-2\)
整数\(a\ne0\),\(b\ne0\),\(m>0\),\(y>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2-abc+abt=ct\),\(a=ct\),\(c\ne\sqrt{k+1}\),\(c\ne m^2-2\)
\(t\ne\sqrt{u+1}\),\(t\ne y^2-2\),整数\(a\ne0\),\(b\ne0\),\(k>0\)
\(m>0\),\(u>0\),\(y>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2-abc+abt=ct\),\(a=ct\),\(c\ne\sqrt{k+1}\),\(t\ne\sqrt{u+1}\)
整数\(a\ne0\),\(b\ne0\),\(k>0\),\(u>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2+abc+abt=ct\),\(a=ct\),\(c\ne m^2-2\),\(t\ne y^2-2\)
整数\(a\ne0\),\(b\ne0\),\(m>0\),\(y>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2+abc-abt=ct\),\(a=ct\),\(c\ne\sqrt{k+1}\),\(c\ne m^2-2\),\(t\ne\sqrt{u+1}\)
\(t\ne y^2-2\),\(c\ne3d\),\(c\ne5f\),\(t\ne3n\),\(t\ne5r\),整数\(a\ne0\),\(b\ne0\),\(d>0\),\(f>0\)
\(k>0\),\(m>0\),\(n>0\),\(r>0\),\(u>0\),\(y>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2+abc-abt=ct\),\(a=ct\),\(c\ne\sqrt{k+1}\),\(t\ne\sqrt{u+1}\)
\(c\ne3d\),\(c\ne5f\),\(t\ne3n\),\(t\ne5r\),整数\(a\ne0\),\(b\ne0\),\(d>0\)
\(f>0\),\(k>0\),\(n>0\),\(r>0\),\(u>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2+abc-abt=ct\),\(a=ct\),\(c\ne m^2-2\),\(t\ne y^2-2\)
\(c\ne3d\),\(c\ne5f\),\(t\ne3n\),\(t\ne5r\),整数\(a\ne0\),\(b\ne0\),\(d>0\)
\(f>0\),\(m>0\),\(n>0\),\(r>0\),\(y>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2-abc+abt=ct\),\(a=ct\),\(c\ne\sqrt{k+1}\),\(c\ne m^2-2\),\(t\ne\sqrt{u+1}\)
\(t\ne y^2-2\),\(c\ne3d\),\(c\ne5f\),\(t\ne3n\),\(t\ne5r\),整数\(a\ne0\),\(b\ne0\),\(d>0\),\(f>0\)
\(k>0\),\(m>0\),\(n>0\),\(r>0\),\(u>0\),\(y>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2-abc+abt=ct\),\(a=ct\),\(c\ne\sqrt{k+1}\),\(t\ne\sqrt{u+1}\)
\(c\ne3d\),\(c\ne5f\),\(t\ne3n\),\(t\ne5r\),整数\(a\ne0\),\(b\ne0\),\(d>0\)
\(f>0\),\(k>0\),\(n>0\),\(r>0\),\(u>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
已知:\(a^2-abc+abt=ct\),\(a=ct\),\(c\ne m^2-2\),\(t\ne y^2-2\)
\(c\ne3d\),\(c\ne5f\),\(t\ne3n\),\(t\ne5r\),整数\(a\ne0\),\(b\ne0\),\(d>0\)
\(f>0\),\(m>0\),\(n>0\),\(r>0\),\(y>0\),奇数\(c>1\),\(t>1\),素数\(p>0\),\(v>0\)
求证:\(c=p\),\(t=v\)
 楼主| 发表于 2024-11-20 14:21 | 显示全部楼层
命题错误,有反例存在
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