$ \frac {a^{4} + a^{3} + 1}{a^{2}b^{2} + ab^{2} + 1}=k\ge 1$ gives $ (a^2+a)(a^2-kb^2)=k-1\ge 0$ if $ k=1$ then $ a=b$ if $ k>1$ then $ k> a^2>a^2>kb^2$ gives $ b^2=0$ then $ a=b$ or $ b=0$

Exactly as solution that I wrote on exam paper

Can you explain the second line please? How did you get the factored form (a*a+a)(a*a-k*b*b) ?

Standard transformation Notice that second line is equivalent to first one.

Already been posted: http://www.mathlinks.ro/viewtopic.php?t=204778

$a^4+a^3+1\vdots a^2b^2+ab^2+1<=>a^4+a^3+1-(a^2b^2+ab^2+1)\vdots a^2b^2+ab^2+1=>a(a^2-b^2)(a+1)\vdots ab^2(a+1)+1=>a=b$

just same as: if $m+n+1|2(m^2+n^2)-1$,then $m=n$

$(a^2b^2+ab^2+1)\le (a^4+a^3+1)$ $a=b$, $b=0$ also a solution $a^2b^2+ab^2+1=1\in\mathbb{Z}$, $P(a,b)\in\mathbb{Z^+}$ Case bash it

Paragdey12 wrote: $(a^2b^2+ab^2+1)\le (a^4+a^3+1)$ how did you get that inequality?