0

Does there exist a natural number $n$ for which there is a supremum of a set $$A = \{a\in\mathbb{Q}^+ | a^3+a\leq n^2\}$$ in the set of rational numbers?

Since $f(a) = a^3+a$ is increasing function for $a\in \mathbb{Q}^+$, set $A$ has upper bound $n^2$, so we need to solve $a^3+a=n^2$? Is there such $n$?

abelian25
  • 49
  • There is no consensus in whether $0$ is a natural number or not, so you should clarify that. – jjagmath Feb 28 '24 at 00:53
  • 1
    $a(a^2+1)$ is not a perfect square for integer $a>0$; $a^2+1$ and $a$ have no common factors and $a^2+1$ is not a perfect square. I think you can show that there's no rational solutions either. – Alex K Feb 28 '24 at 01:00
  • 2
    I dont think its a matter for there being any such $n$ but of there being any such $a$. Let $a = \frac pq;p,q \in \mathbb N; \gcd(p,q) = 1$ then we need $p^3 + pq^2= n^2q^3$ and .... well, we must have $q^3|p^3$ but $\gcd(p,q)=1$ so $q=1$ and we need natural $p=a$ were $a^3+a=n^2$. I think as $a^3+a = a(a^2+1)$ and as $\gcd(a,a^2+1) = 1$ then for $a(a^2+1)$ to be a perfect square we must have both $a$ an $a^2 +1$ be perfect square. But if $a^2$ and $a^2+1$ are both squares the only solution is $a^2 = 0^2$ and $a^2 + 1=1^2$. And thus $a=0$ and $a^3 + a = 0^2$ and $n = 0$. (only case.) – fleablood Feb 28 '24 at 01:26
  • @jjagmath but $0 \not \in \mathbb Q^+$, right? (actually I'm not sure there is consensus on that either). If $a > 0$ then $a^3 + a > 0$ and so we can't have $n = 0$ whether or not $0$ is natural or not. .... but maybe we should specify whether $\mathbb Q^+$ means "non-negative rationals" or "positive rationals" – fleablood Feb 28 '24 at 01:32

0 Answers0