Olympiad MethodsDifficulty 50-100932 tagged problems

Functional Equations

Functional Equations is the habit of determining functions from identities by strategic substitutions and structural constraints. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

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The Key Result

$$\text{The additive Cauchy equation } f(x+y) = f(x) + f(y) \text{ has only } f(x) = cx \text{ under any one of: continuity at a point, monotonicity, or boundedness on an interval.}$$

To pin down an unknown function from an identity, substitute carefully chosen values to extract structural facts (the value at a point, symmetry, injectivity), then verify the candidate solves the original equation.

Why It Works

1
Denote the given identity by $P(x,y)$. Substituting convenient values such as $x=y=0$ often reveals $f(0)$ or a relation the function must obey everywhere.
2
Swapping variables or setting $y=x$ exposes symmetry; comparing $P(x,y)$ with $P(y,x)$ can force properties like $f$ being odd, even, additive, or multiplicative.
3
Use these facts to guess the family of solutions (typically $f(x)=cx$, $f(x)=x+c$, or a constant), since most contest equations have only linear or constant solutions.
4
Crucially, substitution only yields NECESSARY conditions, so finish by plugging the candidate back into the original equation to confirm it actually works — an unverified candidate is not a proof.

Worked Examples

Example 1

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(x+y) = f(x) + f(y)$ for all reals and $f$ is monotonic (Cauchy's equation with a regularity condition).

1
Set $x=y=0$ in $P(x,y)$: $f(0) = f(0) + f(0)$, so $f(0)=0$.
2
By induction on the additivity, $f(nx) = n f(x)$ for every positive integer $n$; combined with $f(0)=0$ and $f(-x) = -f(x)$ (from $P(x,-x)$), this extends to all integers.
3
Writing $x = \frac{p}{q}$ gives $q f\!\left(\frac{p}{q}\right) = f(p) = p f(1)$, so $f(r) = f(1)\,r$ for every rational $r$; let $c = f(1)$.
4
Extend to all reals by a squeeze. Say $f$ is nondecreasing (so $c = f(1) \ge 0$). Fix any real $x$ and pick rationals $r_1 < x < r_2$. Monotonicity gives $f(r_1) \le f(x) \le f(r_2)$, i.e. $c\,r_1 \le f(x) \le c\,r_2$. Letting $r_1 \uparrow x$ and $r_2 \downarrow x$ through rationals, both bounds tend to $cx$, so $f(x) = cx$ (the nonincreasing case is identical with inequalities reversed). Check: $f(x)=cx$ gives $c(x+y)=cx+cy$. ✓

Answer:

$f(x) = cx$ for some constant $c \in \mathbb{R}$.

Contest level

Find all $f:\mathbb{R}\to\mathbb{R}$ with $f(x+y) = f(x) + f(y) + xy$ for all reals.

1
Guess that the $xy$ term comes from a quadratic, so try $f(x) = \tfrac12 x^2 + g(x)$ and see what $g$ must satisfy.
2
Substituting, $\tfrac12(x+y)^2 + g(x+y) = \tfrac12 x^2 + g(x) + \tfrac12 y^2 + g(y) + xy$. The $xy$ from $\tfrac12(x+y)^2 = \tfrac12 x^2 + xy + \tfrac12 y^2$ cancels the $+xy$ on the right.
3
What remains is $g(x+y) = g(x) + g(y)$ — the additive Cauchy equation. Over $\mathbb{Q}$ it forces $g(r) = cr$ with $c = g(1)$; this collapses to $g(x) = cx$ on all of $\mathbb{R}$ once any regularity (continuity at a point, monotonicity, or boundedness on an interval) is imposed, and without such a hypothesis the full solution set is the additive family.
4
Verify $f(x) = \tfrac12 x^2 + cx$: $f(x+y) = \tfrac12(x+y)^2 + c(x+y) = \tfrac12 x^2 + \tfrac12 y^2 + xy + cx + cy = f(x) + f(y) + xy$. ✓

Answer:

$f(x) = \tfrac12 x^2 + cx$ (with $g$ additive in general; $g(x)=cx$ under any mild regularity).

Olympiad / Challenge

Find all $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x^2 - y^2) = (x-y)\bigl(f(x) + f(y)\bigr)$ for all reals.

1
Let $P(x,y)$ be the relation. $P(x,x)$: $f(0) = 0$. $P(x,0)$: $f(x^2) = x\,(f(x) + f(0)) = x f(x)$.
2
$P(0,y)$: $f(-y^2) = (-y)(f(0) + f(y)) = -y f(y)$. Combined with $f(x^2)=x f(x)$, replacing $y$ by $x$ gives $f(-x^2) = -x f(x) = -f(x^2)$, so $f$ is odd.
3
$P(x,1)$: $f(x^2 - 1) = (x-1)(f(x) + f(1))$. Using $f(x^2) = x f(x)$ and $f$ additive on these expressions, expand and compare to deduce $f(x+1) = f(x) + f(1)$, which iterates to additivity; write $c = f(1)$.
4
Additivity together with $f(x^2) = x f(x)$ forces linearity on ALL reals with no regularity assumption: apply $f(x^2) = xf(x)$ to $x+1$ and expand the left side additively, $f((x+1)^2) = f(x^2) + 2f(x) + f(1) = xf(x) + 2f(x) + c$, while the right side is $(x+1)f(x+1) = (x+1)(f(x)+c) = xf(x) + f(x) + xc + c$; equating gives $2f(x) = f(x) + xc$, i.e. $f(x) = cx$ for every real $x$. Verify: $(x-y)(cx + cy) = c(x-y)(x+y) = c(x^2 - y^2) = f(x^2 - y^2)$. ✓

Answer:

$f(x) = cx$ for any constant $c \in \mathbb{R}$.

Going Deeper

Cauchy's equation $f(x+y)=f(x)+f(y)$ has only the linear solutions $f(x)=cx$ once you add ANY regularity (monotone, bounded on an interval, measurable, or continuous at one point); without that, pathological solutions exist via a Hamel basis.
FE problems are an IMO/Putnam staple; the standard kill-chain is $P(0,0)$ for $f(0)$, then injectivity/surjectivity, then a substitution that produces additivity or multiplicativity, then verify.
Pitfall: substitution gives only NECESSARY conditions. Skipping the final check is the most common way to lose points — and a candidate that 'looks forced' can still fail to satisfy the original identity.

Spot the Signal

Look for problems where the key step is determining functions from identities by strategic substitutions and structural constraints.
You can describe the hard part as determining functions from identities by strategic substitutions and structural constraints, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for functional equations, then rewrite the givens around it.
Name the relation that makes Functional Equations legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Functional Equations just because the surface looks familiar; verify the required condition first.
Applying Functional Equations because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let functional equations reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is determining functions from identities by strategic substitutions and structural constraints.

Back to techniques

Olympiad MethodsDifficulty 50-100932 tagged problems

Functional Equations

Practice this technique Match my level

The Key Result

$$\text{The additive Cauchy equation } f(x+y) = f(x) + f(y) \text{ has only } f(x) = cx \text{ under any one of: continuity at a point, monotonicity, or boundedness on an interval.}$$

Why It Works

1
Denote the given identity by $P(x,y)$. Substituting convenient values such as $x=y=0$ often reveals $f(0)$ or a relation the function must obey everywhere.
2
Swapping variables or setting $y=x$ exposes symmetry; comparing $P(x,y)$ with $P(y,x)$ can force properties like $f$ being odd, even, additive, or multiplicative.
3
Use these facts to guess the family of solutions (typically $f(x)=cx$, $f(x)=x+c$, or a constant), since most contest equations have only linear or constant solutions.
4
Crucially, substitution only yields NECESSARY conditions, so finish by plugging the candidate back into the original equation to confirm it actually works — an unverified candidate is not a proof.

Worked Examples

Example 1

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(x+y) = f(x) + f(y)$ for all reals and $f$ is monotonic (Cauchy's equation with a regularity condition).

1
Set $x=y=0$ in $P(x,y)$: $f(0) = f(0) + f(0)$, so $f(0)=0$.
2
By induction on the additivity, $f(nx) = n f(x)$ for every positive integer $n$; combined with $f(0)=0$ and $f(-x) = -f(x)$ (from $P(x,-x)$), this extends to all integers.
3
Writing $x = \frac{p}{q}$ gives $q f\!\left(\frac{p}{q}\right) = f(p) = p f(1)$, so $f(r) = f(1)\,r$ for every rational $r$; let $c = f(1)$.
4
Extend to all reals by a squeeze. Say $f$ is nondecreasing (so $c = f(1) \ge 0$). Fix any real $x$ and pick rationals $r_1 < x < r_2$. Monotonicity gives $f(r_1) \le f(x) \le f(r_2)$, i.e. $c\,r_1 \le f(x) \le c\,r_2$. Letting $r_1 \uparrow x$ and $r_2 \downarrow x$ through rationals, both bounds tend to $cx$, so $f(x) = cx$ (the nonincreasing case is identical with inequalities reversed). Check: $f(x)=cx$ gives $c(x+y)=cx+cy$. ✓

Answer:

$f(x) = cx$ for some constant $c \in \mathbb{R}$.

Contest level

Find all $f:\mathbb{R}\to\mathbb{R}$ with $f(x+y) = f(x) + f(y) + xy$ for all reals.

1
Guess that the $xy$ term comes from a quadratic, so try $f(x) = \tfrac12 x^2 + g(x)$ and see what $g$ must satisfy.
2
Substituting, $\tfrac12(x+y)^2 + g(x+y) = \tfrac12 x^2 + g(x) + \tfrac12 y^2 + g(y) + xy$. The $xy$ from $\tfrac12(x+y)^2 = \tfrac12 x^2 + xy + \tfrac12 y^2$ cancels the $+xy$ on the right.
3
What remains is $g(x+y) = g(x) + g(y)$ — the additive Cauchy equation. Over $\mathbb{Q}$ it forces $g(r) = cr$ with $c = g(1)$; this collapses to $g(x) = cx$ on all of $\mathbb{R}$ once any regularity (continuity at a point, monotonicity, or boundedness on an interval) is imposed, and without such a hypothesis the full solution set is the additive family.
4
Verify $f(x) = \tfrac12 x^2 + cx$: $f(x+y) = \tfrac12(x+y)^2 + c(x+y) = \tfrac12 x^2 + \tfrac12 y^2 + xy + cx + cy = f(x) + f(y) + xy$. ✓

Answer:

$f(x) = \tfrac12 x^2 + cx$ (with $g$ additive in general; $g(x)=cx$ under any mild regularity).

Olympiad / Challenge

Find all $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x^2 - y^2) = (x-y)\bigl(f(x) + f(y)\bigr)$ for all reals.

1
Let $P(x,y)$ be the relation. $P(x,x)$: $f(0) = 0$. $P(x,0)$: $f(x^2) = x\,(f(x) + f(0)) = x f(x)$.
2
$P(0,y)$: $f(-y^2) = (-y)(f(0) + f(y)) = -y f(y)$. Combined with $f(x^2)=x f(x)$, replacing $y$ by $x$ gives $f(-x^2) = -x f(x) = -f(x^2)$, so $f$ is odd.
3
$P(x,1)$: $f(x^2 - 1) = (x-1)(f(x) + f(1))$. Using $f(x^2) = x f(x)$ and $f$ additive on these expressions, expand and compare to deduce $f(x+1) = f(x) + f(1)$, which iterates to additivity; write $c = f(1)$.
4
Additivity together with $f(x^2) = x f(x)$ forces linearity on ALL reals with no regularity assumption: apply $f(x^2) = xf(x)$ to $x+1$ and expand the left side additively, $f((x+1)^2) = f(x^2) + 2f(x) + f(1) = xf(x) + 2f(x) + c$, while the right side is $(x+1)f(x+1) = (x+1)(f(x)+c) = xf(x) + f(x) + xc + c$; equating gives $2f(x) = f(x) + xc$, i.e. $f(x) = cx$ for every real $x$. Verify: $(x-y)(cx + cy) = c(x-y)(x+y) = c(x^2 - y^2) = f(x^2 - y^2)$. ✓

Answer:

$f(x) = cx$ for any constant $c \in \mathbb{R}$.

Going Deeper

Cauchy's equation $f(x+y)=f(x)+f(y)$ has only the linear solutions $f(x)=cx$ once you add ANY regularity (monotone, bounded on an interval, measurable, or continuous at one point); without that, pathological solutions exist via a Hamel basis.
FE problems are an IMO/Putnam staple; the standard kill-chain is $P(0,0)$ for $f(0)$, then injectivity/surjectivity, then a substitution that produces additivity or multiplicativity, then verify.
Pitfall: substitution gives only NECESSARY conditions. Skipping the final check is the most common way to lose points — and a candidate that 'looks forced' can still fail to satisfy the original identity.

Spot the Signal

Look for problems where the key step is determining functions from identities by strategic substitutions and structural constraints.
You can describe the hard part as determining functions from identities by strategic substitutions and structural constraints, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for functional equations, then rewrite the givens around it.
Name the relation that makes Functional Equations legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Functional Equations just because the surface looks familiar; verify the required condition first.
Applying Functional Equations because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let functional equations reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is determining functions from identities by strategic substitutions and structural constraints.