Tìm các số nguyên tố $(u;v)$ sao cho : $(u+v)^{2}|(u^{4}+v^{4})$
Tìm các số nguyên tố $(u;v)$ sao cho : $(u+v)^{2}|(u^{4}+v^{4})$
Started By letankhang, 19-01-2014 - 22:57
#1
Posted 19-01-2014 - 22:57
letankhang
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Posted 19-01-2014 - 23:17
Zaraki
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Tìm các số nguyên tố $(u;v)$ sao cho : $(u+v)^{2}|(u^{4}+v^{4})$
Lời giải. Nếu $u=v$ thì $4u^2|2u^4$ nên $u=v=2$.
Nếu $u \ne v$ thì $\gcd (u,v)=1$. Ta có $u^4+v^4=(u+v)^4-2uv(2u^2+2v^2+3uv)=(u+v)^4-4uv(u+v)^2+2u^2v^2$. Do đó ta suy ra $(u+v)^2|2u^2v^2$ nên $(u+v)^2|2$ (vì $\gcd (u,v)=1$), điều này mâu thuẫn.
Vậy $(u,v)=(2,2)$.
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