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Let $P(x)$ a polynomial in x with real coefficients such that for all real numbers $x, y, z$ satisfying $xy + yz + zx = 1$,$ P(x) + P(y) + P(z) = P(x + y + z)$. Furthermore, $P(0) = 1$ & $P(1) = 4$. Find $P(2017)$.

This looks to me like the Cauchy functional equation which is why the title. I understand that a Cauchy equation has solutions of the type $f(x)=cx$ but can't figure out the case here with that constraint. My wild guess, however, is that $P(x)$ should be a perfect square. Please help.

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  • $\begingroup$ The constraint adds possible solutions. For example, consider $P(x) = x^2+1$. $\endgroup$ – Michael L. Aug 1 '17 at 20:45
  • $\begingroup$ Can you explain a bit more? $\endgroup$ – Epsilon zero Aug 1 '17 at 20:46
  • $\begingroup$ Hint. $P(2)=P(1+1+0)=P(1)+P(1)+P(0)=9$ $\endgroup$ – serg_1 Aug 1 '17 at 20:55
  • $\begingroup$ @AnuranChowdhury $xy+yz+zx = 1$ has the solution $z = \frac{1-xy}{x+y}$. You can check that $$P(x)+P(y)+P\left(\frac{1-xy}{x+y}\right) = P\left(x+y+\frac{1-xy}{x+y}\right)$$ for $P(x) = x^2+1$. This $P$ doesn't solve your problem, of course, as $P(1)\neq 4$. $\endgroup$ – Michael L. Aug 1 '17 at 20:59
  • $\begingroup$ Now that we have one solution $(x+1)^2,$ who can show that there are no others? $\endgroup$ – Reiner Martin Aug 1 '17 at 21:11
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Note that $P(x) = ax^2+bx+a$ solves this problem, as \begin{align*} P(x+y+z)-P(x)-P(y)-P(z) &= a(x+y+z)^2-a(x^2+y^2+z^2)-2a \\ &= 2a(xy+yz+zx-1) \end{align*} Then, we can impose $P(0) = a = 1$ and $P(1) = 2a+b = 4$. The latter implies $b = 2$, so we have $P(x) = x^2+2x+1 = (x+1)^2$. Then we find $P(2017) = 2018^2$.


To prove that this is the only solution, note that for $$P^*(x, y, z) := P(x+y+z)-P(x)-P(y)-P(z)$$ $P^*(x, y, z)$ must be in the ideal generated by $xy+yz+zx-1$ in $\mathbb{C}[x, y, z]$ as this polynomial is irreducible over $\mathbb{C}$. Therefore, $$P^*(x, y, z) = Q(x, y, z)(xy+yz+zx-1)$$ for some polynomial $Q$. Note that $P^*(x, y, z)$ and $xy+yz+zx-1$ are both symmetric polynomials. In particular, $xy+yz+zx-1 = \frac{1}{2}p_1^2-\frac{1}{2}p_2-1$, where $p_k := x^k+y^k+z^k$. Now, note that for $P(x) = \sum_{k=0}^n a_kx^k$, $$P^*(x, y, z) = \sum_{k=0}^n a_k(p_1^k-p_k)$$ As $Q(x, y, z)$ is the quotient of symmetric polynomials, it must also be a symmetric polynomial. Thus, we write $R^*(p_1, p_2, \ldots, p_n)$ and $R(p_1, p_2, \ldots, p_n)$ to be the polynomial representations of $P^*$ and $Q$ respectively in terms of power sums (which are algebraically independent). After switching to $R$ and $R^*$, our equation in $P^*$ and $Q$ becomes $$R^*(p_1, \ldots, p_n) = \left(\frac{1}{2}p_1^2-\frac{1}{2}p_2-1\right)R(p_1, \ldots, p_n)$$ If $a_n\neq 0$ for some $n\geq 3$, then $R^*(p_1, \ldots, p_n)$ will have a nonzero term $-a_np_n$. However, this term cannot appear on the right-hand side, as if $R$ has a term $a_np_n$, then it would imply that $R^*$ also has terms $-\frac{a_n}{2}p_1^2p_n$ and $-\frac{a_n}{2}p_2p_n$ (which do not appear in the above sum). Thus, $R^*$ must equal $-2a_0+a_2(p_1^2-p_2)$, which implies that $\deg(R) = \deg(R^*)-2 = 0$. Then, $R\equiv c$ for some constant $c$, and $a_0 = a_2 = \frac{c}{2}$. Letting $a = \frac{c}{2}$ and $b = a_1$, we have $$P(x) = ax^2+bx+a$$ as above.

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One solution is $P(x)=(x+1)^2,$ as then $$ P(x+y+z)=P(x)+P(y)+P(z)+2(xy+yz+zx-1). $$ Thus, $P(2017)$ can be $2018^2.$

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By $P(2)=P(1+1+0)=P(1)+P(1)+P(0)=9.$ How about assuming $P(x)=(x+1)^2$ Checking $P(x+y+z)=(x+y+z+1)^2=x^2+2x+y^2+2y+z^2+2z+2(xy+yz+xz)+1=x^2+2x+y^2+2y+z^2+2z+1+1+1=(x+1)^2+(y+1)^2+(z+1)^2=P(x)+P(y)+P(z)$

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