\(A=\left(1+b^2+a^2+a^2b^2\right).\left(1+c^2\right)\)
\(=1+a^2+b^2+c^2+a^2c^2+b^2c^2+a^2b^2+a^2b^2c^2\)
\(=1+\left(a+b+c\right)^2-2.\left(ab+bc+ac\right)+\left(ab+bc+ac\right)^2-2abc.\left(a+b+c\right)+a^2b^2c^2\)
Thay ab+bc+ac=1 vào A, ta có:
\(A=1+\left(a+b+c\right)^2-2+1-2abc.\left(a+b+c\right)+a^2b^2c^2\)
\(=\left(a+b+c\right)^2-2abc.\left(a+b+c\right)+a^2b^2c^2\)
\(=\left(a+b+c-abc\right)^2\)
Vì a,b,c thuộc Z
\(\Rightarrow\left(a+b+c-abc\right)^2\)là số chính phương
\(\hept{\begin{cases}\left(1+a^2\right)=\left(ab+bc+ca+a^2\right)=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(a+c\right)\\\left(1+b^2\right)=\left(ab+bc+ca+b^2\right)=a\left(b+c\right)+b\left(b+c\right)=\left(a+b\right)\left(b+c\right)\\\left(1+c^2\right)=\left(ab+bc+ca+c^2\right)=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\end{cases}}\)
\(\Rightarrow A=\text{[}\left(a+b\right)\left(b+c\right)\left(c+a\right)\text{]}^2\Rightarrow\text{đ}pcm\)
