求方程7(x+y)=3(x2-xy+y2)的整数解.
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求方程7(x+y)=3(x2-xy+y2)的整数解.
设x+y=3t,x2-xy+y2=7t(其中t为整数),
∴3xy=(x+y)2-(x2-xy+y2)=9t2-7t,
∴7(x+y)=3(x2-xy+y2)≥3(x2-2xy+y2)=3(x-y)2≥0,
⇒x+y≥0,
∴t≥0,
∵(x-y)2=x2-xy+y2-xy=-
28t−9t2
3,
∴t是3的倍数,于是,设t=3k,
则(x-y)2=28k-27k2=k(28-27k)≥0,
又∵k≥0,
∴k=0或1,
①当k=0时,x+y=0,即x=-y代入x2-xy+y2=0,
解得x=0、y=0;
②当k=1时,则x-y=±1,x+y=3t=9k=9,
解得x=4、y=5或x=5、y=4.
答:原方程的解为x=0、y=0,x=5,y=4或x=4,y=5.
∴3xy=(x+y)2-(x2-xy+y2)=9t2-7t,
∴7(x+y)=3(x2-xy+y2)≥3(x2-2xy+y2)=3(x-y)2≥0,
⇒x+y≥0,
∴t≥0,
∵(x-y)2=x2-xy+y2-xy=-
28t−9t2
3,
∴t是3的倍数,于是,设t=3k,
则(x-y)2=28k-27k2=k(28-27k)≥0,
又∵k≥0,
∴k=0或1,
①当k=0时,x+y=0,即x=-y代入x2-xy+y2=0,
解得x=0、y=0;
②当k=1时,则x-y=±1,x+y=3t=9k=9,
解得x=4、y=5或x=5、y=4.
答:原方程的解为x=0、y=0,x=5,y=4或x=4,y=5.
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