After expanding the term we need to show it's perfect square, we obtain $(abc)^2 + (ab)^2 +(bc)^2 +(ca)^2 +a^2 +b^2 +c^2 +1$, and from $ab+bc+ac=1$ by squaring we obtain: $(ab)^2 +(bc)^2 +(ca)^2=1-2(acb^2+bca^2+abc^2)$ From these two we have: $(abc)^2 + (ab)^2 +(bc)^2 +(ca)^2 +a^2 +b^2 +c^2 +1=(abc)^2 -2(acb^2+bca^2+abc^2)+2+a^2+b^2+c^2$ $\leftrightarrow$ $(abc)^2 -2abc(a+b+c) +a^2+b^2+c^2+2ab+2bc+2ac=(abc)^2 -2abc(a+b+c)+(a+b+c)^2=(abc-a-b-c)^2$ Done

Tomekk wrote: After expanding the term we need to show it's perfect square, we obtain $(abc)^2 + (ab)^2 +(bc)^2 +(ca)^2 +a^2 +b^2 +c^2 +1$, and from $ab+bc+ac=1$ by squaring we obtain: $(ab)^2 +(bc)^2 +(ca)^2=1-2(acb^2+bca^2+abc^2)$ From these two we have: $(abc)^2 + (ab)^2 +(bc)^2 +(ca)^2 +a^2 +b^2 +c^2 +1=(abc)^2 -2(acb^2+bca^2+abc^2)+2+a^2+b^2+c^2$ $\leftrightarrow$ $(abc)^2 -2abc(a+b+c) +a^2+b^2+c^2+2ab+2bc+2ac=(abc)^2 -2abc(a+b+c)+(a+b+c)^2=(abc-a-b-c)^2$ Done It's too long. Just observe $a^2+1=(a+b)(c+a)$.

\[(1+a^2)(1+b^2)(1+c^2)=(a+i)(b+i)(c+i)(a-i)(b-i)(c-i)\]\[=((abc-a-b-c)+(ab+bc+ca-1)i)((abc-a-b-c)-(ab+bc+ca-1)i)=(abc-a-b-c)^2\]

a^2 + 1 = a^2 + ab + bc + ca = (a+b)(a+c) (a^2+1)(b^2+1)(c^2+1) = (a+b)^2 * (b+c)^2 * (c+a)^2

Which integers satisfy $ab+bc+ca=1$?

@above to name a few: $(1,0,1), (-1,0,-1), (-3, 5,8)$