Difference between revisions of "2016 USAMO Problems/Problem 4"

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'''Step 7:''' By Step 6 and the second observation from Step 4, the only possible solutions are <math>f \equiv 0</math> and <math>f(x) = x^2</math> for all <math>x \in \mathbb{R}.</math> It's easy to check that both of these work, so we're done.
 
'''Step 7:''' By Step 6 and the second observation from Step 4, the only possible solutions are <math>f \equiv 0</math> and <math>f(x) = x^2</math> for all <math>x \in \mathbb{R}.</math> It's easy to check that both of these work, so we're done.
 
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==Solution 2==
 
==Solution 2==
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Finally, substitute <math>y=-x</math>. Then <math>(f(x)-x^2)f(4x)+(f(y)-x^2)f(4x)=0</math>, so <math>2(f(x)-x^2)f(4x)=0</math>. If <math>f(x)-x^2=0, \boxed{f(x)=x^2}</math>. If <math>f(4x)=0</math>, we've already found the solution. Substitution of <math>f</math> for both cases works, and because <math>f</math> has to apply for all real numbers, we are done.
 
Finally, substitute <math>y=-x</math>. Then <math>(f(x)-x^2)f(4x)+(f(y)-x^2)f(4x)=0</math>, so <math>2(f(x)-x^2)f(4x)=0</math>. If <math>f(x)-x^2=0, \boxed{f(x)=x^2}</math>. If <math>f(4x)=0</math>, we've already found the solution. Substitution of <math>f</math> for both cases works, and because <math>f</math> has to apply for all real numbers, we are done.
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==See also==
 
==See also==

Revision as of 18:14, 16 April 2018

Problem

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, \[(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.\]

Solution 1

Step 1: Set $x = y = 0$ to obtain $f(0) = 0.$

Step 2: Set $x = 0$ to obtain $f(y)f(-y) = f(y)^2.$

$\indent$ In particular, if $f(y) \ne 0$ then $f(y) = f(-y).$

$\indent$ In addition, replacing $y \to -t$, it follows that $f(t) = 0 \implies f(-t) = 0$ for all $t \in \mathbb{R}.$

Step 3: Set $x = 3y$ to obtain $\left[f(y) + 3y^2\right]f(8y) = f(4y)^2.$

$\indent$ In particular, replacing $y \to t/8$, it follows that $f(t) = 0 \implies f(t/2) = 0$ for all $t \in \mathbb{R}.$

Step 4: Set $y = -x$ to obtain $f(4x)\left[f(x) + f(-x) - 2x^2\right] = 0.$

$\indent$ In particular, if $f(x) \ne 0$, then $f(4x) \ne 0$ by the observation from Step 3, because $f(4x) = 0 \implies f(2x) = 0 \implies f(x) = 0.$ Hence, the above equation implies that $2x^2 = f(x) + f(-x) = 2f(x)$, where the last step follows from the first observation from Step 2.

$\indent$ Therefore, either $f(x) = 0$ or $f(x) = x^2$ for each $x.$

$\indent$ Looking back on the equation from Step 3, it follows that $f(y) + 3y^2 \ne 0$ for any nonzero $y.$ Therefore, replacing $y \to t/4$ in this equation, it follows that $f(t) = 0 \implies f(2t) = 0.$

Step 5: If $f(a) = f(b) = 0$, then $f(b - a) = 0.$

$\indent$ This follows by choosing $x, y$ such that $x - 3y = a$ and $3x - y = b.$ Then $x + y = \tfrac{b - a}{2}$, so plugging $x, y$ into the given equation, we deduce that $f\left(\tfrac{b - a}{2}\right) = 0.$ Therefore, by the third observation from Step 4, we obtain $f(b - a) = 0$, as desired.

Step 6: If $f \not\equiv 0$, then $f(t) = 0 \implies t = 0.$

$\indent$ Suppose by way of contradiction that there exists an nonzero $t$ with $f(t) = 0.$ Choose $x, y$ such that $f(x) \ne 0$ and $x + y = t.$ The following three facts are crucial:

$\indent$ 1. $f(y) \ne 0.$ This is because $(x + y) - y = x$, so by Step 5, $f(y) = 0 \implies f(x) = 0$, impossible.

$\indent$ 2. $f(x - 3y) \ne 0.$ This is because $(x + y) - (x - 3y) = 4y$, so by Step 5 and the observation from Step 3, $f(x - 3y) = 0 \implies f(4y) = 0 \implies f(2y) = 0 \implies f(y) = 0$, impossible.

$\indent$ 3. $f(3x - y) \ne 0.$ This is because by the second observation from Step 2, $f(3x - y) = 0 \implies f(y - 3x) = 0.$ Then because $(x + y) - (y - 3x) = 4x$, Step 5 together with the observation from Step 3 yield $f(3x - y) = 0 \implies f(4x) = 0 \implies f(2x) = 0 \implies f(x) = 0$, impossible.

$\indent$ By the second observation from Step 4, these three facts imply that $f(y) = y^2$ and $f(x - 3y) = \left(x - 3y\right)^2$ and $f(3x - y) = \left(3x - y\right)^2.$ By plugging into the given equation, it follows that \begin{align*} \left(x^2 + xy\right)\left(x - 3y\right)^2 + \left(y^2 + xy\right)\left(3x - y\right)^2 = 0. \end{align*} But the above expression miraculously factors into $\left(x + y\right)^4$! This is clearly a contradiction, since $t = x + y \ne 0$ by assumption. This completes Step 6.

Step 7: By Step 6 and the second observation from Step 4, the only possible solutions are $f \equiv 0$ and $f(x) = x^2$ for all $x \in \mathbb{R}.$ It's easy to check that both of these work, so we're done.

Solution 2

Step 1: $x=y=0 \implies f(0)=0$

Step 2: $x=0 \implies f(y)f(-y)=f(y)^{2}$. Now, assume $y \not = 0$. Then, if $f(y)=0$, we substitute in $-y$ to get $f(y)f(-y)=f(-y)^{2}$, or $f(y)=f(-y)=0$. Otherwise, we divide both sides by $f(y)$ to get $f(y)=f(-y)$. If $y=0$, we obviously have $f(0)=f(0)$. Thus, the function is even. . Step 3: $y=-x \implies 2f(4x)(f(x)-x^{2})=0$. Thus, $\forall x$, we have $f(4x)=0$ or $f(x)=x^{2}$.

Step 4: We now assume $f(x) \not = 0$, $x\not = 0$. We have $f(\frac{x}{4})=\frac{x^{2}}{16}$. Now, setting $x=y=\frac{x}{4}$, we have $f(\frac{x}{2}=\frac{x^{2}}{4}$ or $f(\frac{x}{2})=0$. The former implies that $f(x)=0$ or $x^{2}$. The latter implies that $f(x)=0$ or $f(x)=\frac{x^{2}}{2}$. Assume the latter. $y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}$. Clearly, this implies that $f(x)$ is negative for some $m$. Now, we have $f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0$, which is a contradiction. Thus, $\forall x$$f(x)=0$ or $f(x)=x^{2}$.

Step 5: We now assume $f(x)=0$, $f(y)=y^{2}$ for some $x,y \not = 0$. Let $m$ be sufficiently large integer, let $z=|4^{m}x|$ and take the absolute value of $y$(since the function is even). Choose $c$ such that $3z-c=y$. Note that we have $\frac{c}{z}$~$3$ and $\frac{y}{z}$~$0$. Note that $f(z)=0$. Now, $x=z, y=c \implies$ LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to $(z+c)^{4}$~$256z^{4}$. Now if $f(z-3c)=0$, the second term of the LHS/RHS clearly ~0 as $m \to \infty$. if $f(z-3c)=0$, then we have LHS/RHS ~ $0$, otherwise, we have LHS/RHS~$\frac{8^{2}\cdot 3z^{4}}{256z^{4}}$~$\frac{3}{4}$, a contradiction, as we're clearly not dividing by $0$, and we should have LHS/RHS=1.

Solution Simple Substitution (3)

Each of these steps is rather intuitive. Obviously the function can only be solved by substitution.

Substitute $x=0, y=0$ to get $f(0)^2+f(0)^2=f(0)^2.$ Clearly $f(0)^2=0, f(0)=0$.

Substitute $x=0$ only to get $f(y)f(-y)=f(y)^2$. Either $f(y)=0, f(-y)=f(y)$. In the first case a solution is $\boxed{f(x)=0}$. In the second, we know $f$ is even.

Finally, substitute $y=-x$. Then $(f(x)-x^2)f(4x)+(f(y)-x^2)f(4x)=0$, so $2(f(x)-x^2)f(4x)=0$. If $f(x)-x^2=0, \boxed{f(x)=x^2}$. If $f(4x)=0$, we've already found the solution. Substitution of $f$ for both cases works, and because $f$ has to apply for all real numbers, we are done.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2016 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6
All USAMO Problems and Solutions
2016 USAJMO (ProblemsResources)
Preceded by
Problem 5
Last Problem
1 2 3 4 5 6
All USAJMO Problems and Solutions
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