Let $p(x)$ and $q(x)\neq 0$ be polynomials with coefficients in $\mathbb{Q}$. If $\frac{p(\pi)}{q(\pi)}=\frac{p(\pi+n)}{q(\pi+n)}$ for all $n\in\mathbb{Z}$, then can we conclude that $\frac{p(\pi)}{q(\pi)}$ must be a rational?
Asked
Active
Viewed 74 times
2
-
1Yes. As $\pi$ is transcendental over $\Bbb{Q}$ we have that $p(x)=q(x)$ if and only if $p(\pi)=q(\pi)$. Therefore the answer to your previous question is also an answer to this question. – Jyrki Lahtonen Apr 09 '19 at 06:25
-
In other words, I think this is a duplicate. However, I see a little bit of rooom for disagreement, so I won't use my (immediately binding) dupehammer without a voice of support. – Jyrki Lahtonen Apr 09 '19 at 06:27
-
1I guess I should've asked the previous question about $p,q$ being polynomials with rational coefficients.. my mistake. But ok – user500144 Apr 09 '19 at 06:28
2 Answers
2
$p(\pi)q(x+\pi)-q(\pi)p(x+\pi)$ vanishes at all integer points. Since it is a polynomial it must vanish identically. Put $x =-\pi$ to get $p(\pi)q(0)=q(\pi)p(0)$. Hence $\frac {p(\pi)} {q(\pi)}=\frac {p(0)} {q(0)}$ which is rational.
Kavi Rama Murthy
- 359,332
-
By "since it is a polynomial it must vanish identically", you mean that since a polynomial can only have finitely many zeroes and all integers are zeroes for this expression, this expression is the identically zero polynomial? – user500144 Apr 09 '19 at 06:36
-
2
The rational function $$ \frac{p(x)}{q(x)}-\frac{p(\pi)}{q(\pi)} $$ (a priori with real coefficients, not necessarily rational) has infinitely many zeroes. That means it must be constantly $0$. So $\frac{p(x)}{q(x)}$ must also be a constant, which clearly has to be a rational number.
Arthur
- 204,511