# How to solve this 2017 USATST problem at last problem

(2017 USATST) Let $P, Q \in \mathbb{R}[x]$ be relatively prime nonconstant polynomials. Show that there can be at most three real numbers $\lambda$ such that $P + \lambda Q$ is the square of a polynomial.

Assume otherwise, then there exist four $\lambda_1,\lambda_2,\lambda_3,\lambda_4$ such that $P+\lambda_i Q=R_i^2$. Differentiating, we get that $P'+\lambda_i Q'=2R_i R_i'$. Therefore, $R_i$ divides both $P+\lambda_i Q$ and $P'+\lambda_i Q'$, so it divides $$Q*(P'+\lambda_i Q')-Q'*(P+\lambda_i Q)=P' Q-Q'P.$$

Note that since $P$ and $Q$ are coprime, we clearly must also have $(Q,R_i)=(P,R_i)=1$. If $R_1$ and $R_2$ share some root $c$, then $P(c)+\lambda_1 Q(c)=P(c)+\lambda_2 Q(c)=0$. Since $Q(c) \neq 0$, we get $\lambda_1=\lambda_2$, contradiction. Therefore the $R_i$ are pairwise coprime, so by CRT we get $R_1R_2R_3R_4 \mid P' Q-Q'P$

If $P$ and $Q$ have different degrees, then $\deg(P' Q-Q'P) \leq \deg(P)+\deg(Q)-1$, but \begin{align} \deg(R_1 R_2R_3R_4) &= \deg(R_1)+\deg(R_2)+\deg(R_3)+\deg(R_4) \\ &=4 \frac{\max(\deg(P),\deg(Q))}2 \\ &=2 \max(\deg(P),\deg(Q)) \end{align}

This would imply that $P'Q-Q'P=0$ as a polynomial, but invoking the quotient rule(!), we get that $\bigg(\frac{P}Q\bigg)'=0$, which would imply that either $P$ and $Q$ are not relatively prime, or they share a common factor.

If $P$ and $Q$ have the same degree.How to solve it?

• You can't assume that $(Q,R_i)=(P,R_i)=1$, because $\lambda_i$ might be $0$. – TonyK Apr 13 '17 at 13:05
• We only need $R_i$ are pairwise coprime. If $(R_1,R_2)=D$, then from $\begin{cases} P+\lambda_1Q=R_1^2\\ P+\lambda_2Q=R_2^2\end{cases}$, we know D divides both $P$ and $Q$. @TonyK – C.Ding Apr 13 '17 at 13:34
• @C.Ding: Yes. But the OP says "we clearly must also have $(Q,R_i)=(P,R_i)=1$", which is false. – TonyK Apr 13 '17 at 13:46
• I wonder is there three $\lambda$s which make $P + \lambda Q$ perfect square. All I can think of is $(x \pm 1)^2$... – guest Apr 17 '17 at 7:10

You can assume that $P,Q$ have different degrees: if they have the same degree then there is some $c$ such that $\deg(P-cQ) < \deg P$, so you can write $P + \lambda_i Q = (P-cQ) + (\lambda_i + c) Q$ and proceed as in your solution with $P$ replaced by $P-cQ$.
You must still contend with the possibility that $P$ has smaller degree but one of the $\lambda_i$ is zero. This makes the corresponding $R_i$ degree smaller as well. Fortunately $\deg (P'Q - PQ')$ then drops as well and the argument still goes through.
One can prove more: it's not even possible for $\prod_{i=1}^4 (P+\lambda_i Q)$ to be a square with pairwise distinct $\lambda_i$ unless $P=cQ$ (or $Q=0$). Indeed if $R^2 = \prod_{i=1}^4 (P+\lambda_i Q)$ then we have a map from the projective line to the elliptic curve $r^2 = \prod_{i=1}^4 (p+\lambda_i q)$, and such a map must be constant, so $P:Q$ is constant. This kind of argument may be out of bounds for USATST, but it should be possible to unwind it to something like the argument you gave (one proof of the result about maps from ${\bf P}^1$ to an elliptic curve uses the Riemann-Hurwitz formula, which can be proved using degrees of differentials, which comes down to the degree of the numerator of $(P/Q)'$).
• I don't understand what you mean by "$\deg(P'Q-PQ')$ drops as well" ? – Ewan Delanoy Apr 19 '17 at 9:36
• The bound $\deg P + \deg Q - 1$ decreases compared with the $2\max(\deg P, \deg Q)$ that appears in the OP's partial solution. – Noam D. Elkies Apr 19 '17 at 17:44
• @NoamD.Elkies :Can you show the details of the case that one of the $\lambda_i$ is zero. I can't work out. The case is important, since the '$c$' you give must be one of the $-\lambda_i$. – C.Ding Apr 20 '17 at 11:58