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Đặt \(p=2k+1\)( phụ chú : vì p là số nguyên tố lẻ )
\(x=a-b-c\)
\(y=b-c-a\)
\(z=c-a-b\)
\(\Rightarrow-\left(x+y+z\right)=a+b+c\)
\(\Rightarrow B=x^{2k+1}+y^{2k+1}+z^{2k+1}-\left(x+y+z\right)^{2k+1}\)
\(=\left(x^{2k+1}+y^{2k+1}\right)-\left[\left(x+y+z\right)^{2k+1}-z^{2k+1}\right]\)
\(=\left(x+y\right)\left(x^{2k}-x^{2k-1}y+....+y^{2k}\right)-\left(x+y\right)\left[\left(x+y+z\right)^{2k}+\left(x+y+z\right)^{2k-1}z+...+z^{2k}\right]\)chia hết cho \(x+y=-2c\)
\(\Rightarrow B\text{⋮}c\)
Tiếp, lại có :
\(B=x^{2k+1}+y^{2k+1}+z^{2k+1}-\left(x+y+z\right)^{2k+1}\)
\(=\left(x^{2k+1}+z^{2k+1}\right)-\left[\left(x+y+z\right)^{2k+1}-y^{2k+1}\right]\)
\(=\left(x+z\right)\left(x^{2k}-x^{2k-1}z+...+z^{2k}\right)-\left(x+z\right)\left[\left(x+y+z\right)^{2k}+\left(x+y+z\right)^{2k-1}y+...+y^{2k}\right]\)chia hết cho \(x+z=-2b\)
\(\Rightarrow B\text{⋮}b\)
CMTT, có \(B\text{⋮}a\)
Mà \(a,b,c\)đôi một nguyên tố cùng nhau ( GT )
\(\Rightarrow B\text{⋮}abc\)
Vậy ...
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Ta có:
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\Leftrightarrow\left(a+b\right)c=ab\Leftrightarrow ab-bc-ab=0\)
Hay \(ab-bc-ab+c^2=c^2\Leftrightarrow\left(b-c\right)\left(a-c\right)=c^2\)
Nếu \(\left(b-c;a-c\right)=d\ne1\Rightarrow c^2=d^2\left(loai\right)\)
Vậy \(\left(b-c;a-c\right)=1\Rightarrow c-b;c-a\) là 2 số chính phương
Đặt \(b-c=n^2;a-c=m^2\)
\(\Rightarrow a+b=b-c+a-c+2c=m^2+n^2+2mn=\left(m+n\right)^2\) là số chính phương