The evenness bottleneck in a quadratic functional equation

bidoof_supreme·Jan 22, 2026

The candidates f=0 and f(x)=x² work. My problem is proving f is even: comparing y and −y seems to give two alternatives that could change with y. Problem: Find all functions

f: \mathbb{R} \to \mathbb{R}

satisfying, for all

x \neq 0

and all

y

f(x + y^2) = f(x) + f(y)^2 + \frac{2f(xy)}{x}.

Source: Baltic Way 2015 Shortlisted Problems · MNET-01CF

Open MNET-01CF in the question bank →

6 replies

no_mp_left·Jan 22, 2026

For fixed y, the comparison says

f(y)^2-f(-y)^2+2(f(xy)-f(-xy))/x=0

. At x=1, either f(y)=f(−y) or f(y)+f(−y)=−2.

bidoof_supreme·Jan 22, 2026

So I can't choose the same alternative for every y. But if the equality alternative holds for even one nonzero y, the comparison for arbitrary x forces global evenness.

no_mp_left·Jan 23, 2026

Exactly. If it never holds for nonzero y, the sum alternative holds everywhere off zero. The comparison becomes

f(xy)+1=x(f(y)+1)

, so f(x)=ax−1 for x≠0. Substitution back gives

(a^2-a)y^2+1=2/x

whenever the arguments are nonzero, impossible with y=1 and x=1,2.

suspicious_barrel·Jan 23, 2026

With evenness established, set y=1 and also replace x by −x−1. The two equations are

f(x+1)=(1+2/x)f(x)+f(1)^2

and

f(x)=(1-2/(x+1))f(x+1)+f(1)^2

. Eliminate f(x+1) to obtain

f(x)=x^2f(1)^2

away from x=0,−1.

bidoof_supreme·Jan 23, 2026

I still need those excluded arguments. y=0 in the original equation, for two choices of nonzero x, forces f(0)=0. Evenness handles −1, and x=1 in the formula gives f(1)=f(1)².

no_mp_left·Jan 23, 2026

Then f(1) is 0 or 1 and the formula covers all reals. Direct substitution verifies both solutions, completing the classification.

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The evenness bottleneck in a quadratic functional equation

bidoof_supreme·Jan 22, 2026

The candidates f=0 and f(x)=x² work. My problem is proving f is even: comparing y and −y seems to give two alternatives that could change with y. Problem: Find all functions

f: \mathbb{R} \to \mathbb{R}

satisfying, for all

x \neq 0

and all

y

f(x + y^2) = f(x) + f(y)^2 + \frac{2f(xy)}{x}.

Source: Baltic Way 2015 Shortlisted Problems · MNET-01CF

Open MNET-01CF in the question bank →

6 replies

no_mp_left·Jan 22, 2026

For fixed y, the comparison says

f(y)^2-f(-y)^2+2(f(xy)-f(-xy))/x=0

. At x=1, either f(y)=f(−y) or f(y)+f(−y)=−2.

bidoof_supreme·Jan 22, 2026

So I can't choose the same alternative for every y. But if the equality alternative holds for even one nonzero y, the comparison for arbitrary x forces global evenness.

no_mp_left·Jan 23, 2026

Exactly. If it never holds for nonzero y, the sum alternative holds everywhere off zero. The comparison becomes

f(xy)+1=x(f(y)+1)

, so f(x)=ax−1 for x≠0. Substitution back gives

(a^2-a)y^2+1=2/x

whenever the arguments are nonzero, impossible with y=1 and x=1,2.

suspicious_barrel·Jan 23, 2026

With evenness established, set y=1 and also replace x by −x−1. The two equations are

f(x+1)=(1+2/x)f(x)+f(1)^2

and

f(x)=(1-2/(x+1))f(x+1)+f(1)^2

. Eliminate f(x+1) to obtain

f(x)=x^2f(1)^2

away from x=0,−1.

bidoof_supreme·Jan 23, 2026

I still need those excluded arguments. y=0 in the original equation, for two choices of nonzero x, forces f(0)=0. Evenness handles −1, and x=1 in the formula gives f(1)=f(1)².

no_mp_left·Jan 23, 2026

Then f(1) is 0 or 1 and the formula covers all reals. Direct substitution verifies both solutions, completing the classification.