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 $$\deg(R_1 R_2R_3R_4)=\deg(R_1)+\deg(R_2)+\deg(R_3)+\deg(R_4)=4 \frac{\max(\deg(P),\deg(Q))}2=2\max(\deg(P),\deg(Q))$$ 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, we have to be a bit more careful, but we can still get that $\deg(P' Q-Q'P) \leq\deg(R_1 R_2R_3R_4)$ and conclude in the same way.

The way to take of the case where $\deg P=\deg Q$ relies on the fact that $\gcd(P+cQ,Q)=\gcd(P,Q)=1$ for any $c\in\mathbb R$. This basically says that we can shift $P$ by any multiple of $Q$ without affecting the problem, so we may shift $P$ so that $\deg P\ne \deg Q$ and finish as above.

Does anyone know a solution which does not rely on calculus?

mathcool2009 wrote: Does anyone know a solution which does not rely on calculus? Here's an outline of the official solution. This is true even with ${\mathbb R}$ replaced by ${\mathbb C}$, and it will be necessary to work in this generality. We will prove the problem in the following form. Problem: Assume $P$ and $Q$ are relatively prime in $\mathbb C[x]$. If $\alpha P + \beta Q$ is a square for four different choices of the ratio $[\alpha : \beta]$ then $P$ and $Q$ must be constant. Call pairs $(P,Q)$ as in the claim bad; so we wish to show the only bad pairs are pairs of constant polynomials. Assume not, and take a bad pair with $\deg P + \deg Q$ minimal. By a suitable Moebius transformation, we may transform $(P,Q)$ so that the four ratios are $[1:0]$, $[0:1]$, $[1,-1]$ and $[1,-k]$, so we find there are polynomials $A$ and $B$ such that \begin{align*} A^2 - B^2 &= C^2 \\ A^2 - k B^2 &= D^2 \end{align*}where $P = A^2$ and $Q = B^2$ (after the transformation). Consequently, we have $C^2 = (A+B)(A-B)$ and $D^2 = (A+\mu B)(A-\mu B)$ where $\mu^2 = k$. Now $\gcd(A,B) = 1$, so $A+B$, $A-B$, $A+ \mu B$ and $A - \mu B$ are squares; id est $(A,B)$ is bad. This is a contradiction, since $\deg A + \deg B < \deg P + \deg Q$.

This is a useful lemma for IOM 2016 Q5.

Please correct me if I am wrong. Lemma: If Q is a non-constant real polynomial, then there are at most one a in R for which Q + a is the square of a polynomial. Proof: If Q+ a_i =R_i^2 for i=1,2 then( R1+R2)(R1-R2) = a1-a2. So R1+R2 and R1-R2 are both constants. Hence R1 and R2 and so Q is constant. Contradiction. Now come to the main problem. We shall prove the stronger statement with 'three' replaced by 'two'. Suppose a_i , i=1,2,3 are three such lambdas. As P and Q are relatively prime, so there are c,d in R with cP+dQ=1. As P and Q are non-constant, so c,d are nonzero. So, (1/c)+[a_i-(d/c)]Q=R_i^2. At most one of a_i-(d/c) can be zero. WLOG asume a_1 and a_2 are non zero. a_i-(d/c) for i=1,2 has same sign e as that of the leading coeff. of Q. Dividing by the numerical value, we see that there are two reals g_i[=(c|a_i-(d/c)|)^-1] such that eQ+g_i is a square of a polynomial. This is impossible by the lemma. Contradiction.

I am sorry, for asking this here, but I see amazing knowledge of polynomials, so does anyone know any good resource for learning them(to the level seen in this problem...)

Supravat wrote: Please correct me if I am wrong. Lemma: If Q is a non-constant real polynomial, then there are at most one a in R for which Q + a is the square of a polynomial. Proof: If Q+ a_i =R_i^2 for i=1,2 then( R1+R2)(R1-R2) = a1-a2. So R1+R2 and R1-R2 are both constants. Hence R1 and R2 and so Q is constant. Contradiction. Now come to the main problem. We shall prove the stronger statement with 'three' replaced by 'two'. Suppose a_i , i=1,2,3 are three such lambdas. As P and Q are relatively prime, so there are c,d in R with cP+dQ=1. As P and Q are non-constant, so c,d are nonzero. So, (1/c)+[a_i-(d/c)]Q=R_i^2. At most one of a_i-(d/c) can be zero. WLOG asume a_1 and a_2 are non zero. a_i-(d/c) for i=1,2 has same sign e as that of the leading coeff. of Q. Dividing by the numerical value, we see that there are two reals g_i[=(c|a_i-(d/c)|)^-1] such that eQ+g_i is a square of a polynomial. This is impossible by the lemma. Contradiction. c and d are polynomials. You can't take inverses of polynomials.

How can this very well-known, very old problem have been given for a USA TST? I mean, common people, are you running out of ideas ? I first saw this problem in the first lecture of my Elliptic Curves course(in fairness, that's way after I finished high-school ). Also,how could "Alison Miller" have signed the problem or been indicated as the author? Please

Replace $\mathbb{R}$ with any field $K$ of characteristic $\ne 2,3$. WLOG $K=\overline{K}$. Assume for contradiction there exist $P,Q\in K[x]$ with $4$ such $\lambda$. We may let one of the $\lambda$ be $0$, so let the 4 distinct values be $0,-\frac{1}{\lambda_1},-\frac{1}{\lambda_2},-\frac{1}{\lambda_3}$. Let $R^2=P(Q-\lambda_1P)(Q-\lambda_2P)(Q-\lambda_3P)$, $R\in K[x]$ since $P-\frac{Q}{\lambda}$ square means $-\lambda(P-\frac{Q}{\lambda})=Q-\lambda P$ square. Let $\varphi=\frac{R}{P^2},\psi=\frac{Q}{P}$. Then we have that \[\varphi^2=(\psi-\lambda_1)(\psi-\lambda_2)(\psi-\lambda_3).\]Consider the elliptic curve $E/K$ given by $y^2=(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$, which is non-singular since it has distinct roots. Then $(x,y)=(\varphi,\psi)$ is a rational parametrization of $E/K$, which is clearly false because elliptic curves are not rational. Thus we obtain our desired contradiction.

@v_Enhance could you tell us more about how Möbius transformations work on polynomials and what properties they preserve?

Could you tell us why it has distinct roots?

zacchro wrote: 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 $$\deg(R_1 R_2R_3R_4)=\deg(R_1)+\deg(R_2)+\deg(R_3)+\deg(R_4)=4 \frac{\max(\deg(P),\deg(Q))}2=2\max(\deg(P),\deg(Q))$$ 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, we have to be a bit more careful, but we can still get that $\deg(P' Q-Q'P) \leq\deg(R_1 R_2R_3R_4)$ and conclude in the same way. Why $\deg(R_i)=\frac{\deg(P)+\deg(Q)}{2}$ in fact there exist a unique $\lambda$ to let $\deg(P+\lambda Q)$ is lower.

artsolver wrote: I am sorry, for asking this here, but I see amazing knowledge of polynomials, so does anyone know any good resource for learning them(to the level seen in this problem...) You can see Yufei Zhao's notes on integer polynomials which was used at MOP !

liekkas wrote: zacchro wrote: 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 $$\deg(R_1 R_2R_3R_4)=\deg(R_1)+\deg(R_2)+\deg(R_3)+\deg(R_4)=4 \frac{\max(\deg(P),\deg(Q))}2=2\max(\deg(P),\deg(Q))$$ 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, we have to be a bit more careful, but we can still get that $\deg(P' Q-Q'P) \leq\deg(R_1 R_2R_3R_4)$ and conclude in the same way. Why $\deg(R_i)=\frac{\deg(P)+\deg(Q)}{2}$ in fact there exist a unique $\lambda$ to let $\deg(P+\lambda Q)$ is lower. Supposedly $2deg(R_i)=\max \{\deg(P),\deg(Q)\}\ge \deg(P)+\deg(Q) $, if leading coefficients of $ P, \lambda Q$ are not equal.

The following solution is essentially v_Enhance's solution presented in the way I thought about. Hopefully this is more accessible version. We relax the problem's condition slightly by replacing $\mathbb{R}[x]$ with $\mathbb{C}[x]$. Assume that there exists $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ such that $P+\lambda_j Q = R_j^2$ for each $j=1,2,3,4$. As $P,Q$ are relatively prime, $R_1,R_2,R_3,R_4$ are pairwise relatively prime. Claim: There exists relatively prime polynomials $f,g$ and constant $k$ such that $f, g, f-g, f-kg$ are perfect square. Proof: By linear algebra, vectors $(1,\lambda_1)$, $(1,\lambda_2)$ and $(1,\lambda_3)$ are linearly dependent. Therefore there exists constants $k_1, k_2, k_3$ such that $$k_1\cdot R_1^2 + k_2\cdot R_2^2 + k_3\cdot R_3^2 = 0$$Let $f = k_1R_1^2$, $g = -k_2R_2^2$, we get $f,g,f-g$ are perfect square. Finally, using the existence of $\lambda_4$, we get vector $(1,\lambda_4)$ must be a linear combination of $(1,\lambda_1)$ and $(1,\lambda_2)$. This gives $\alpha f+\beta g$ is a perfect square for some $\alpha, \beta \ne 0$. And we are done by setting $k = -\tfrac{\beta}{\alpha}$. Seeing this claim, we ignore $P$ and $Q$ and prove the following. Claim: There are no relatively prime polynomials $f,g\in\mathbb{C}[x]$ and $k\in\mathbb{C}$ such that $f, g, f-g, f-kg$ are all perfect square. Proof: Let $f = A^2$, $g=B^2$ and $\mu = \sqrt{k}$. Then as $(A-B)(A+B)$ is perfect square, we get $f_1 = (1+\mu)(A+B)$ and $g_1 = (1-\mu)(A-B)$ are perfect square too. Repeating the similar logic, we get $f _1- g_1 = 2(A + \mu B)$ and $f_1 +c_1g_1 = c_2(A+\mu B)$ are perfect square for some constants $c_1,c_2$. Thus we are done by infinite descent on $(f,g)\to (f_1,g_1)$.

Solved with goodbear and Th3Numb3rThr33. Assume for contradiction $R_i^2=P+\lambda_iQ$ for $i=1,2,3,4$. Note that \[\gcd\big(R_i^2,R_j^2\big)=\gcd\big(P+\lambda_iQ,Q(\lambda_i,\lambda_j)\big)=\gcd\big(P,Q(\lambda_i-\lambda_j)\big)\mid \lambda_i-\lambda_j,\]which is constant. Thus $R_i$ and $R_j$ are all relatively prime. We will assume $\deg P\ne\deg Q$; otherwise shift $P$ by an appropriate multiple of $Q$ so that the leading coefficient vanishes. Differentiating, we obtain $2R_iR_i'=P'+\lambda_iQ'$, so \[R_i\mid Q(P'+\lambda_iQ')-Q'(P+\lambda_iQ)=QP'-Q'P=Q^2(P/Q)'.\]This polynomial is not $0$ since $\gcd(P,Q)=1$, and since $R_i$ are relatively prime, by Chinese Remainder theorem \[R_1R_2R_3R_4\mid Q^2(P/Q)'.\]Since $\deg P\ne \deg Q$, we have $\deg R_i\le\tfrac12\max(\deg P,\deg Q)$; equality holds for at least $3$ of them, and if equality does not hold, then $\lambda_i=0$ and $\deg R_i=\tfrac12\min(\deg P,\deg Q)$. Hence \begin{align*} \deg(R_1R_2R_3R_4)&\ge\tfrac32\max(\deg P,\deg Q)+\tfrac12\min(\deg P,\deg Q)\\ &>\deg P+\deg Q-1=\deg\left[Q^2(P/Q)'\right], \end{align*}contradiction.

v_Enhance wrote: By a suitable Moebius transformation, we may transform $(P,Q)$ so that the four ratios are $[1:0]$, $[0:1]$, $[1,-1]$ and $[1,-k]$. Please explain (or cite some links) as to what IS a MOBIUS TRANSFORMATION and how does it help us here..

Bump! Anyone who knows?

Solved with rama1728. Very nice! We prove via contradiction. We work in $\mathbb{C}.$ Take $P,Q$ not both constant with $\max(\deg P, \deg Q)$ minimal. Suppose $P + \lambda_1Q = R_1^2,P + \lambda_2Q = R_2^2,P + \lambda_3Q = R_3^2,P + \lambda_4Q = R_4^2$ for polynomials $R_1,R_2,R_3,R_4.$ It's then easy to see both $P$ and $Q$ are nonconstant. Since $P,Q$ are relatively prime, $R_1,R_2,R_3,R_4$ are pairwise relatively prime. Since the $\lambda$ are all distinct, there are reals $a,b$ where $aR_1^2 + bR_2^2 = R_3^2$ and reals $c,d$ where $c/d \ne a/b$ and $cR_1^2 + dR_2^2 = R_4^2.$ If we set $A = aR_1^2$ and $B=bR_2^2,$ then we know at least one of these is nonconstant. Also, the polynomials $A,B, A-B$ are squares and there exists a constant $k \ne 1$ where $A-kB$ is a square as well. Let $A= S^2, B = T^2$ for polynomials $S,T.$ Note $S$ and $T$ are relatively prime, at least one of them is nonconstant, and $\deg S, \deg T < \max(\deg P, \deg Q).$ Also $(S+T)(S-T) = A-B$ and $(S+T\sqrt{k})(S-T\sqrt{k}) = A-kB$ are squares. So $S+T,S-T, S+T\sqrt{k}, S-T\sqrt{k}$ are squares. But then $S,T$ can be our $P,Q;$ contradiction. $\blacksquare$

WLOG $P = R^2, P+Q = (R+M)^2, P+kQ = (R+N)^2$ are three solutions. Then $Q = (2R+M)M, kQ = (2R+N)N$. So let $2R + M = AC, M = BD, 2R + N = AD, N = kBC$. Then by SFFT, $(A+B)(C-D) = (1-k)BC$. Now notice if any two of $A, B, C, D$ share a nonconstant factor, that factor will divide $Q$ and by extension from the cases it will divide $R$, which is a contradiction since $gcd(P = R^2, Q) = 1$. Hence they are pairwise coprime, so for real constants $m, n$ such that $\frac{m}{n} = k$ we have $A+B = \frac{1}{n}C, C-D = (n-m)B$ so $C = nA + nB, D = nA + mB$. Using these values, $2R = AC - BD = nA^2 - mB^2$ and $Q = ABCD = AB(nA+nB)(nA+mB)$ so $(N + 2R)M = Q$. Thus $N$ is unique given a choice of $M$ and fixed $Q, R$, so there cannot be a fourth distinct solution which can replace $N$.

WLOG $deg(P) \neq deg(Q)$ as we can shift $P$ by a multiple of $Q$, FTSOC assume that there exist $\lambda_1 \cdots \lambda_4$ so that $P + \lambda_i Q = {R_i}^2$, differentiate to get $P'+\lambda_i Q' = 2R_i R'_i$ so $R_i \mid P' + \lambda_i Q'$ and $R_i \mid P + \lambda_i Q$ and therefore $R_i \mid P'Q-Q'P$ Claim 1: $R_i$ are all coprime

and therefore , we have $R_1R_2R_3R_4 \mid P'Q-Q'P$ however the degree of LHS is more than degree of RHS, hence $P'Q=Q'P$ however this implies that $P$ and $Q$ have a common root, contradiction! and we're done.

Quick remark ! As the statetment can also be easily proved in $\mathbb{C}$ , This gives an alternate proof of fermarts last theorom for polynomials when $n=4$ for if $f^4+g^4=h^4$ where $(f,g)=1$. and non constant This forces $(f^2+ig^2)(f^2-ig^2)=h^4$ or $(f+e^{\frac{\pi}{2}i}g)(f+e^{\frac{\pi}{2}i 2 }g)(f+e^{\frac{\pi}{2}i3}g)(f+e^{\frac{\pi}{2}i4}g)=h^4$ As all terms in the product are relatively prime as well we know each of them is perfect square , which by this problem is not possible