Overview

A polynomial in several variables is symmetric if it is unchanged under any permutation of its variables. For example, r2+s2r^2 + s^2 is symmetric (swapping gives s2+r2s^2 + r^2, which is identical), but r+2sr + 2s is not (swapping gives s+2r≠r+2ss + 2r \neq r + 2s). Symmetric polynomials arise naturally from polynomial roots: every expression Vieta's formulas produce is symmetric in the roots by construction. The basic idea is that since we are able to switch up various variables without changing it, we can use various constructions to our advantage

Elementary Symmetric Polynomials

The elementary symmetric polynomials SkS_k in variables x1,…,xnx_1, \ldots, x_n are defined as the sum of all distinct products of exactly kk variables:

S1=∑ixi,S2=∑i<jxixj,S3=∑i<j<kxixjxk,…,Sn=x1x2⋯xnS_1 = \sum_i x_i, \qquad S_2 = \sum_{i < j} x_i x_j, \qquad S_3 = \sum_{i < j < k} x_i x_j x_k, \quad \ldots, \quad S_n = x_1 x_2 \cdots x_n

For three variables x,y,zx, y, z explicitly:

S1=x+y+z,S2=xy+yz+zx,S3=xyzS_1 = x+y+z, \qquad S_2 = xy+yz+zx, \qquad S_3 = xyz

The number of terms in SkS_k is (nk)\binom{n}{k}, since each term chooses kk variables from nn.

Generating function interpretation. The elementary symmetric polynomials are the coefficients in the expansion

∏i=1n(1+xit)=1+S1t+S2t2+⋯+Sntn\prod_{i=1}^n (1 + x_i t) = 1 + S_1 t + S_2 t^2 + \cdots + S_n t^n

Fundamental Theorem of Symmetric Polynomials

Theorem. Every symmetric polynomial with integer (or rational, or real) coefficients can be written uniquely as a polynomial in S1,S2,…,SnS_1, S_2, \ldots, S_n.

Key Techniques

The ss–pp Substitution (Two Variables)

For two variables x,yx, y, set s=x+ys = x + y and p=xyp = xy. Then:

ExpressionIn terms of s,ps, p
x2+y2x^2 + y^2s2−2ps^2 - 2p
x3+y3x^3 + y^3s3−3sp=s(s2−3p)s^3 - 3sp = s(s^2 - 3p)
x4+y4x^4 + y^4s4−4s2p+2p2s^4 - 4s^2p + 2p^2
x2y+xy2x^2 y + xy^2spsp
(x−y)2(x-y)^2s2−4ps^2 - 4p

The key is (x+y)k(x+y)^k expands with cross terms that must be subtracted; never guess the formula, always expand.

For three variables, the analogous substitution is u=x+y+zu = x+y+z, v=xy+yz+zxv = xy+yz+zx, w=xyzw = xyz.

Palindromic (Even Symmetric) Polynomials

For a degree-2n2n palindromic polynomial such as ax4+bx3+cx2+bx+a=0ax^4 + bx^3 + cx^2 + bx + a = 0, divide by xnx^n and substitute y=x+x−1y = x + x^{-1}:

x2+x−2=y2−2,x3+x−3=y3−3y,xk+x−k=Tk(y)x^2 + x^{-2} = y^2 - 2, \qquad x^3 + x^{-3} = y^3 - 3y, \qquad x^k + x^{-k} = T_k(y)

where TkT_k satisfies the same recurrence Tk=y⋅Tk−1−Tk−2T_k = y \cdot T_{k-1} - T_{k-2}. This reduces a degree-2n2n polynomial to a degree-nn polynomial in yy.

Constructing Polynomials with Shifted Roots

To evaluate (r1+c)(r2+c)⋯(rn+c)(r_1 + c)(r_2 + c) \cdots (r_n + c) where rir_i are roots of f(x)f(x): substitute x=y−cx = y - c to get a new polynomial g(y)=f(y−c)g(y) = f(y-c) whose roots are ri+cr_i + c. The product equals ±g(0)/an\pm g(0)/a_n by Vieta's.

Worked Examples

Example 1. Express x2+y2+z2x^2 + y^2 + z^2 in terms of S1,S2,S3S_1, S_2, S_3.

Expand S12=(x+y+z)2=x2+y2+z2+2(xy+yz+zx)=x2+y2+z2+2S2S_1^2 = (x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx) = x^2+y^2+z^2 + 2S_2.

Rearranging: x2+y2+z2=S12−2S2\boxed{x^2+y^2+z^2 = S_1^2 - 2S_2}.

Example 2. Given a+b=1a+b = 1 and a2+b2=2a^2+b^2 = 2, find a3+b3a^3+b^3.

Set s=a+b=1s = a+b = 1 and p=abp = ab. From the sum of squares:

a2+b2=s2−2p  ⟹  2=1−2p  ⟹  p=−12a^2+b^2 = s^2-2p \implies 2 = 1-2p \implies p = -\tfrac{1}{2}

Now apply the sum of cubes identity:

a3+b3=s(s2−3p)=1 ⁣⋅ ⁣(1−3 ⁣⋅ ⁣(−12))=1⋅52=52a^3+b^3 = s(s^2-3p) = 1\!\cdot\!\left(1 - 3\!\cdot\!\left(-\tfrac{1}{2}\right)\right) = 1 \cdot \tfrac{5}{2} = \boxed{\tfrac{5}{2}}

Example 3. Let a,b,ca, b, c be roots of f(x)=x3+8x+5=0f(x) = x^3 + 8x + 5 = 0. Find (1+a)(1+b)(1+c)(1+a)(1+b)(1+c).

Do not use SkS_k directly — instead construct a polynomial with roots 1+a,1+b,1+c1+a, 1+b, 1+c. Substitute x=y−1x = y - 1:

g(y)=(y−1)3+8(y−1)+5=y3−3y2+3y−1+8y−8+5=y3−3y2+11y−4g(y) = (y-1)^3 + 8(y-1) + 5 = y^3 - 3y^2 + 3y - 1 + 8y - 8 + 5 = y^3 - 3y^2 + 11y - 4

By Vieta's for gg, the product of its roots is −(−4)/1=4-(-4)/1 = \boxed{4}.

Example 4. Solve x4+x3+x2+x+1=0x^4 + x^3 + x^2 + x + 1 = 0.

The coefficients are symmetric (palindromic). Divide by x2x^2:

x2+x+1+x−1+x−2=0x^2 + x + 1 + x^{-1} + x^{-2} = 0

Group: (x2+x−2)+(x+x−1)+1=0(x^2 + x^{-2}) + (x + x^{-1}) + 1 = 0. Let u=x+x−1u = x + x^{-1}, so x2+x−2=u2−2x^2 + x^{-2} = u^2 - 2:

(u2−2)+u+1=0  ⟹  u2+u−1=0  ⟹  u=−1±52(u^2-2) + u + 1 = 0 \implies u^2 + u - 1 = 0 \implies u = \frac{-1 \pm \sqrt{5}}{2}

For each value of uu, solve x+x−1=ux + x^{-1} = u, i.e., x2−ux+1=0x^2 - ux + 1 = 0, giving the four roots of the original.

Example 5 (Discriminant). Determine whether a polynomial has a repeated root using only its coefficients.

A polynomial f(x)f(x) with roots x1,…,xnx_1, \ldots, x_n has a repeated root iff Δ=∏i<j(xi−xj)2=0\Delta = \prod_{i < j}(x_i - x_j)^2 = 0. Since Δ\Delta is symmetric in the roots, the Fundamental Theorem guarantees it is a polynomial in S1,…,SnS_1, \ldots, S_n. By Vieta's, each SkS_k is a coefficient of ff up to sign. Therefore Δ\Delta — the discriminant — is computable purely from the coefficients of ff using arithmetic operations.

Common Pitfalls

Asymmetric expressions. An expression must be unchanged under every variable permutation, not just one specific swap. r+2sr + 2s is not symmetric and cannot be expressed in SkS_k alone.

Sign errors in Vieta's. S1=−an−1/anS_1 = -a_{n-1}/a_n (negative when an−1>0a_{n-1} > 0); S2=+an−2/anS_2 = +a_{n-2}/a_n (positive); signs alternate. This is the most common arithmetic error.

Wrong ss–pp expansions. x3+y3≠s3x^3+y^3 \neq s^3. The correct identity is x3+y3=s3−3spx^3+y^3 = s^3 - 3sp. Always expand carefully.

Palindromic substitution cross-term. When substituting y=x+x−1y = x + x^{-1}, do not write x2+x−2=y2x^2 + x^{-2} = y^2. The correct identity is x2+x−2=y2−2x^2 + x^{-2} = y^2 - 2 (the cross term 2⋅x⋅x−1=22 \cdot x \cdot x^{-1} = 2 must be subtracted).

Root multiplicity. If every root of g(x)g(x) is also a root of f(x)f(x), this does not mean g∣fg \mid f. You must verify that the multiplicity of each shared root in gg does not exceed its multiplicity in ff.

Practice Problems

StatusSourceProblem NameDifficultyTags
AIMEEasy
Show TagsSymmetric Polynomial, Systems of Equations, Vieta's Formulas
AMC 12Hard
Show TagsLogarithms, Sum of Cubes, Symmetric Polynomial
AIMEHard
Show TagsNewton's Identities, Recurrence Relations, Symmetric Polynomial
Berkeley Math Circle Monthly Contest 5Hard
Show TagsConstructions and loci, Symmetric functions, Triangle inequalities
Berkeley Math Circle Monthly Contest 5Hard
Show TagsPolynomial operations, Symmetric functions, Vieta's formulas
Berkeley Math Circle: Monthly Contest 4Hard
Show TagsIntegers, Linear and quadratic inequalities, Polynomial operations, Symmetric functions
Berkeley Math Circle Monthly ContestHard
Show TagsPolynomial operations, Symmetric functions
Berkeley Math Circle Monthly Contest 1Hard
Show TagsPolynomial operations, Symmetric functions
Berkeley Math Circle: Monthly Contest 7Hard
Show TagsCauchy-Schwarz, Symmetric functions
Berkeley Math Circle Monthly Contest 1Hard
Show TagsComplex numbers, Recurrence relations, Symmetric functions, Vieta's formulas
Berkeley Math CircleHard
Show TagsPolynomial operations, Symmetric functions, Vieta's formulas
Harvard-MIT Math TournamentHard
Show TagsPolynomial operations, Symmetric functions
Harvard-MIT Mathematics TournamentHard
Show TagsPolynomial operations, Symmetric functions
Harvard-MIT Mathematics TournamentHard
Show TagsSymmetric functions, Vieta's formulas
USA IMO 2003Hard
Show TagsCauchy-Schwarz, Linear and quadratic inequalities, QM-AM-GM-HM / Power Mean, Symmetric functions
USA IMO 2003Hard
Show TagsSymmetric functions
Harvard-MIT Mathematics TournamentHard
Show TagsSymmetric functions, Vieta's formulas
Harvard-MIT Mathematics TournamentHard
Show TagsSymmetric functions, Vieta's formulas
IMOHard
Show TagsCauchy-Schwarz, QM-AM-GM-HM / Power Mean, Symmetric functions
Harvard-MIT Mathematics TournamentHard
Show TagsSymmetric functions
Harvard-MIT Mathematics Tournament, Team Round BHard
Show TagsSymmetric functions
Harvard-MIT Mathematics TournamentHard
Show TagsIntermediate Value Theorem, Recurrence relations, Symmetric functions, Vieta's formulas
Harvard-MIT Mathematics TournamentHard
Show TagsIntermediate Value Theorem, Symmetric functions, Vieta's formulas
$10^{\text {th }}$ Annual Harvard-MIT Mathematics TournamentHard
Show TagsSymmetric functions, Vieta's formulas
Harvard-MIT Mathematics TournamentHard
Show TagsQM-AM-GM-HM / Power Mean, Symmetric functions, Vieta's formulas

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