Show that there is no polynomial P(x) of degree 998 with real coefficients satisfying the equation $P(x)^2 - 1 = P(x^2 + 1)$ for all real numbers x.

Question

A friend of mine explained this solution to me but there are a few parts which I still don't understand. Her solution is as follows:

By substituting in $x$ and $-x$ into the equation given, it can be seen that $P(x)^2 - 1 = P(x^2 + 1) = P((-x)^2 + 1) = P(-x)^2 - 1$, and so it is implied that $P(x) = P(-x)$ or $-P(-x)$. Assume there is a contradiction and there is a polynomial of degree 998 satisfying the equation, then $P(x)$ has an even degree of 998 and so it must also be an even polynomial (1. Why is this the case?), that is $P(x)=Q(x^2)$ for some polynomial Q with degree 499.

By setting $x^2=y$ and using the definition of the polynomial Q, it can be seen that for $y \geq 0$, $Q(y)^2 -1 = Q(y^2 + 2y +1)$. By substituting $y$ and $-2-y$ (2. Why these two?), we get $Q(y) = Q(-2-y)$ or $-Q(-2-y)$.

Now set $R(y) = Q(y-1)$ (3. Why?), then since R has an odd degree of 499 (4. Why?) it can only be an odd function so $R(y)=-R(-y)$. By substituting in $y=0$, this gives $R(0) = -R(0)$ and so $R(0)=0$. Therefore $Q(-1)=0$. Computing values of Q for 1, 4, 25, etc. we get $Q(1)=-1, Q(4)=0, Q(25)=-1,...$ (5. How do we get these?) and so Q has infinitely many zeros which is a contradiction.

Any help explaining the 5 questions inset into this solution will be much appreciated.

Thanks

Answer 1

1. For any polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ with $a_n\ne 0$, the leading term $a_nx^n$ dominates as $|x|\to\infty$. In particular, $P(x)$ and $P(-x)$ have the same sign for $|x|\gg0$ if $n$ is even. Hence at least for $|x|\gg0$, we can only have $P(x)=P(-x)$ and not $P(x)=-P(-x)$. But that means that the polynomials $P(x)$ and $P(-x)$ agree in infinitely many points. This is only possible if they are completely equal, i.e., $P(x)=P(-x)$ for all $x$.

2. Because they work. Perhaps it is easier to see once you notice that $y^2+2y+1=(y+1)^2$. So if we write $z$ for $y+1$, we have $Q(z-1)^2-1=Q(z^2)$. The right hand side is the same for $-z$ as is for $z$, hence $Q(-z-1)^2-1=Q((-z)^2)=Q(z^2)=Q(z-1)^2-1$ and so $Q(-z-1)=\pm Q(z-1)$. Translating back, this says $Q(-y-2)=\pm Q(y)$.

3. I think it was already visible in my answer to $2$, that $z$ is "nicer" than $y$. In $Q(-z-1)=\pm Q(z-1)$, we can get rid of the $-1$ precisely by considering $R(z):=Q(z-1)$ and thereby obtain $R(-z)=\pm R(z)$.

4. As $R(x)=Q(x-1)$, we have $\deg R=\deg Q$. As $P(x)=Q(x^2)$, we have $\deg P=2\deg Q$.

5. If we know $Q(y)$, we also know $Q((y+1)^2)=Q(y)^2-1$. So from $Q(-1)=R(0)=0$, $Q(0)=Q(-1)^2-1=-1$, $Q(1)=Q(0)^2-1=0$, $Q(4)=Q(1)^2-1=-1$, and so on. More formally, we can show by induction that $Q((2k-1)^2)=0$ for all $k$.