Let $a,b$, and $c$ be odd positive integers such that $a$ is not a perfect square and $$a^2+a+1 = 3(b^2+b+1)(c^2+c+1).$$Prove that at least one of the numbers $b^2+b+1$ and $c^2+c+1$ is composite.
Problem
Source: 2019 Baltic Way P18
Tags: number theory
MarkBcc168
18.11.2019 14:26
Just follows the solution of the rather known problem with $a^2+1=(b^2+1)(c^2+1)$.
WLOG $b\geq c$ and let $p=b^2+b+1$. Thus $p\mid (a^2+a+1)-(b^2+b+1) = (a-b)(a+b+1)$ which means $p\mid a-b$ or $p\mid a+b+1$. Consider the following cases.
Case 1: $a-b = p$
This gives $a=b^2+2b+1=(b+1)^2$, contradiction.
Case 2: $a+b+1=p$
This gives $a=b^2$, contradiction.
Case 3: $a+b+1\geq 2p \implies a\geqq 2b^2+b+1$
Since $b\geq c$, we get
$$(2b^2+b+1)^2+(2b^2+b+1)+1 \leq 3(b^2+b+1)^2$$which expands to $b(b-3)(b^2+b+1)\leq 0$. Thus $b\leq 3$.
If $b=1$, then $c=1$ which means $a^2+a+1=27$, contradiction.
If $b=3$, then either $c=1$ or $c=3$. This means $a^2+a+1\in\{117,507\}$ which yields $a=22$, contradiction to $a$ odd.
AKS_9_54_61
18.11.2019 17:40
You can post these in your blog
grupyorum
18.11.2019 19:53
Mark, you can clean up your inequalities further: Assume $b\geqslant c$. Play the same game, and note that if $a+b+1=2p$ then since $b$ is odd, this yields $a$ is even, which contradicts with the preamble. Hence, $a+b+1\geqslant 3p=3(b^2+b+1)\Rightarrow a>2(b^2+b+1)$. Now, $a^2+a+1>a^2>4(b^2+b+1)^2$. On the other hand, $a^2+a+1=3(b^2+b+1)(c^2+c+1)\leqslant 3(b^2+b+1)^2<4(b^2+b+1)^2$. These two signal the desired contradiction immediately. It seems to me that the facts that $a,b,c$ are odd and $a$ is not a perfect square are solely given to simplify the case work above.
Samusasuke
21.08.2020 02:04
The condition that $a$ isn't a perfect square is unnecessary.