Find all positive integers $ a$ and $ b$ such that $ \frac{a^{4}+a^{3}+1}{a^{2}b^{2}+ab^{2}+1}$ is an integer.
Problem
Source: Federation of Bosnia, 3. and 4. Grades 2008.
Tags: number theory proposed, number theory
24.04.2008 00:28
$ \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$
24.04.2008 00:37
Exactly as solution that I wrote on exam paper
04.06.2008 01:30
Can you explain the second line please? How did you get the factored form (a*a+a)(a*a-k*b*b) ?
04.06.2008 20:17
Standard transformation Notice that second line is equivalent to first one.
04.06.2008 20:53
Already been posted: http://www.mathlinks.ro/viewtopic.php?t=204778
25.05.2010 11:02
$a^4+a^3+1\vdots a^2b^2+ab^2+1<=>a^4+a^3+1-(a^2b^2+ab^2+1)\vdots a^2b^2+ab^2+1=>a(a^2-b^2)(a+1)\vdots ab^2(a+1)+1=>a=b$
25.05.2010 14:29
just same as: if $m+n+1|2(m^2+n^2)-1$,then $m=n$
02.02.2018 15:45
$(a^2b^2+ab^2+1)\le (a^4+a^3+1)$ $a=b$, $b=0$ also a solution $a^2b^2+ab^2+1=1\in\mathbb{Z}$, $P(a,b)\in\mathbb{Z^+}$ Case bash it
15.05.2020 22:45
Paragdey12 wrote: $(a^2b^2+ab^2+1)\le (a^4+a^3+1)$ how did you get that inequality?