Find all polynomials $p(x)$ with real coefficients and $x\ p(x+1) = p(x+2)$.
So if we substitute $x = 0$ we get $p(2) = 0$.
If we substitute $x = 1$ we get $p(3) = 0$ and if we continue we can say for $n > 1 : p(n) = 0$.
But what about $n \leq 1$?
Asked
Active
Viewed 70 times
0
-
How do we get $p(3)=0$ if we substitute $x=1$? – Another User Jul 14 '23 at 21:38
-
3Recall that a polynomial of degre $n$ has at most $n$ roots. – Anne Bauval Jul 14 '23 at 21:38
-
5Something is off. The constant $p(x)=0$ is the only solution. For otherwise the polynomial on the left has a higher degree than the one on the right. – Jyrki Lahtonen Jul 14 '23 at 21:45
-
For the "correct" version the techniques from here and/or here will possibly help. – Jyrki Lahtonen Jul 14 '23 at 21:47
-
@JyrkiLahtonen exactly – farnood gholampoor Jul 14 '23 at 21:47
-
My previous comment meant: you proved that $p$ has infinitely many roots. Hence it cannot have a degree $n\in\Bbb Z_{\ge0}$. Therefore, $p=0.$ – Anne Bauval Jul 15 '23 at 21:40
1 Answers
3
You could prove by induction that $p(n)=0$ for all $n\ge2$: the base case is $p(2)=0p(1)$; if $p(n)=0$, then $p(n+1)=p(n)(n-1)=0$. Then use the well-know result that a polynomial with infinitely many roots must be $0$.
A simpler approach is to observe that $p(x+2)$ and $p(x+1)$ have the same degree, therefore if $p\ne 0$ then $\deg(p(x+2))=\deg(xp(x+1))=\deg(p(x+1))+1$, absurd.