Olympiad Algebra IV: Symmetric Functions & the Pointwise Trap

That theorem is the licence for the whole method: the moment an expression is symmetric in the roots, it is computable from the coefficients without ever finding a root. If $a+b+c = 6$, $ab+bc+ca = 11$, $abc = 6$, then $a^3+b^3+c^3 = e_1^3 - 3e_1e_2 + 3e_3 = 216 - 198 + 18 = 36$ — and indeed the numbers are $1, 2, 3$.

For $x^3 - 5x^2 - 8x + 12$: $e_1 = 5$, $e_2 = -8$, $e_3 = -12$, so $p_2 = 25 + 16 = 41$ and $p_3 = 205 + 40 - 36 = 209$. The practical version needs no memorised identity at all — every root satisfies $r^3 = 5r^2 + 8r - 12$, so summing gives $p_3 = 5p_2 + 8p_1 - 36$, and the same trick pushes any $p_k$ down to known ones. Use the recursion the polynomial itself hands you; quote Newton when you need the general shape.

For $p$ of degree $\le 5$ with $p(k) = k^2$ at $k = 1,\dots,5$ and $p(6) = 0$: set $q = p - x^2$, so $q = c(x-1)\cdots(x-5)$; then $q(6) = -36 = 120c$ gives $c = -\tfrac{3}{10}$ and $p(7) = 49 - \tfrac{3}{10}\cdot 720 = -167$. Lagrange interpolation would get the same answer with six times the arithmetic; finite differences are the third route, and are the fastest when the given values are consecutive.

Step 1 — let the other roots be $c,d$ and put $s=a+b$, $p=ab$, $t=c+d$, $q=cd$. Then $x^4+x^3-1=(x^2-sx+p)(x^2-tx+q)$, and comparing coefficients: $s+t=-1$, $p+q+st=0$, $pt+qs=0$, $pq=-1$.

Step 2 — eliminate: $q=-\tfrac1p$ and $t=-1-s$. The third equation becomes $-p(1+s)-\tfrac sp=0$, so $s\left(p+\tfrac1p\right)=-p$ and $s=-\dfrac{p^2}{p^2+1}$.

Step 3 — the second equation becomes $p-\tfrac1p-s-s^2=0$. Substitute $s$ and multiply by $p(p^2+1)^2$:

So $p=ab$ is a root of $x^6+x^4+x^3-x^2-1$. (Here $p\ne0$ since $pq=-1$, and $p^2+1\ne0$ because otherwise $p+\tfrac1p=0$ and the third equation would give $p=0$.)

How to find it. A statement about a pair of roots is a statement about a quadratic factor. Writing the quartic as a product of two quadratics turns it into four equations in four unknowns, and the target polynomial is what remains after eliminating everything except $p$.

Vieta's formulas $e_1=-\tfrac{a_{n-1}}{a_n}$, $e_2=\tfrac{a_{n-2}}{a_n}$, $e_3=-\tfrac{a_{n-3}}{a_n}$ divide by the leading coefficient. Here $e_1=3$, $e_2=\tfrac52$, $e_3=\tfrac72$, so $r^2+s^2+t^2=e_1^2-2e_2=9-5=4$. Make the polynomial monic before reading off symmetric functions; every identity in this section assumes it.

The counterexample is one line: $f(x) = x$ for rational $x$ and $f(x) = -x$ otherwise satisfies $f(x)^2 = x^2$ everywhere and is neither function. So a pointwise conclusion must be followed by a GLUING argument: assume $f(a) = a$ and $f(b) = -b$ with $a, b \ne 0$, feed the pair back into the original equation, and derive a contradiction. With additivity, $f(a+b) = a - b$ must equal $\pm(a+b)$, forcing $a = 0$ or $b = 0$ — so exactly two functions survive. This is the single most common way to lose points on an olympiad functional equation.

So the hypotheses in the problem statement are the solution's scaffolding: if the problem says monotone, the intended path is Cauchy plus monotonicity. The standard substitutions that reduce an equation to Cauchy are worth drilling: $g(x) = f(x) - \tfrac{x^2}{2}$ turns $f(x+y) = f(x)+f(y)+xy$ into Cauchy, giving $f(x) = \tfrac{x^2+x}{2}$ when $f(1) = 1$, so $f(10) = 55$.

A good write-up says what each substitution BUYS: "setting $y = 0$ gives $f(0) = 0$", "so $f$ is injective, hence the $c$ found above is unique". Keep a list of established facts as you go, and end by verifying that every candidate function actually satisfies the original equation — an olympiad answer is incomplete without the check, and graders deduct for its absence.

Each root satisfies $r^3 = e_1r^2 - e_2r + e_3$ (its own defining cubic). Summing over the three roots gives $p_3 = e_1p_2 - e_2p_1 + 3e_3$, the $3$ appearing because the constant $e_3$ is added once per root. $\blacksquare$ The same argument for a degree-$n$ polynomial gives $p_k$ for $k \ge n$; the identities for $k < n$ need the combinatorial proof, which is why the recursion from the polynomial is the practical tool.

Suppose $f(a) = a$ and $f(b) = -b$ with $a, b \ne 0$. Additivity gives $f(a+b) = a - b$, while the pointwise condition gives $f(a+b) = \pm(a+b)$. The plus case forces $b = 0$, the minus case forces $a = 0$ — both contradictions. So the sign is the same at every nonzero point, and both signs do give solutions. $\blacksquare$ Significance: the contradiction used the equation at a NEW point, $a+b$. That is always the shape of the gluing step.