When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

We establish a few initial conditions: $f(0^2-0^2) = f(0) = 0$. $f(x^2-(-x)^2) = f(0) = xf(x)+xf(-x) = 0 \Rightarrow f(x) = -f(-x)$ (1) $f(x^2-0) = f(x^2) = xf(x)$ (2). Now let $a,b$ be positive reals. Then $f(a-b) = \sqrt{a}f(\sqrt{a})+\sqrt{b}f(\sqrt{b})$, which by (2) equals $f(a-b) = f(a)-f(b)$. $f(a-b)+f(b) = f(a)$ (3). Using (1), we can derive the equations $f(a-b)+f(-a) = f(-b)$ (4) $f(-a)+f(b) = f(b-a)$ (5). The following casework is necessary because $a,b$ are positive reals while $x,y$ can be positive or negative. From (3), we set $a = x+y$, $b = y$ to get $f(x)+f(y) = f(x+y)$ for $x,y \ge 0$. From (4), we set $a = -y$, $b = -(x+y)$ to get $f(x)+f(y) = f(x+y)$ for $x,y \le 0$. From (5), we set $a = -x$, $b = y$ OR $a = -y$, $b = x$ to get $f(x)+f(y) = f(x+y)$ for one of $x,y$ negative and the other positive. Hence we have $f(x)+f(y) = f(x+y)$ for all real $x,y$. So $f(x) = f(1)+f(1)+\cdots+f(1) = xf(1)$. Since $f(1)$ is constant, let $f(1) = k$. Then we have $f(x) = kx$ for all real $k$. QED.

paladin8 wrote: So $f(x) = f(1)+f(1)+\cdots+f(1) = xf(1)$. Since $f(1)$ is constant, let $f(1) = k$. Then we have $f(x) = kx$ for all real $k$. QED. so um you've proven it for integers x... what if x isn't an integer?

paladin8, a typo: (1) should read $f(x)=-f(-x)$. ThAzN1's objection also stands. The Cauchy equation $f(x)+f(y) = f(x+y)$ which you have derived does not imply $f(x)=kx$ for all $x$, unless you can also establish that $f(x)$ is continuous, or monotonic, or bounded on some finite interval.

I'll prove that $f(x)=xf(1)$ for every real x. Let $x\in \mathbb{R}$. $f(x)=f(x\cdot 1)$ Looking for $a$ and $b$ such that $x=a+b$ and $1=a-b$ we find $a=\frac{x+1}{2},\ b=\frac{x-1}{2}$, thus $f(x)=f((a+b)(a-b))=f(a^2-b^2)=af(a)-bf(b)=\frac{x+1}{2}f(\frac{x+1}{2})-\frac{x-1}{2}f(\frac{x-1}{2})$. Since $f(t+u)=f(t)+f(u)$ (see Paladin8's solution) we have $f(t/2+t/2)=f(t/2)+f(t/2)$, hence $f(t/2)=f(t)/2$. We get $f(x)=\frac{x+1}{2}\cdot \frac{f(x)+f(1)}{2}-\frac{x-1}{2}\cdot \frac{f(x)-f(1)}{2}$, so $f(x)=\frac{xf(1)+f(x)}{2}$, hence $f(x)=xf(1)$ for every real x.

I guess this is more or less the same thing, but After establishing that $f(x)$ is odd and the Cauchy thing... we need only consider $x \geq 0$. Note that $f(2x+1) = f((x+1)^2-x^2)$ $= (x+1)f(x+1) - xf(x) = f(x+1) + x(f(x+1)-f(x))$ $= f(x+1)+ xf(1)$ Subtracting yields $f(x) = f(2x+1) - f(x+1) = xf(1)$ Replace $x$ with $-x$ and using $f(x) = - f(-x)$ you get the result...

we immediately obtain $f(0)=0$ and we know $f(x-y)=f(x)-f(y)$ let $x+y=1$ then $f(x-y)=xf(x)-yf(y)$ $\Rightarrow f(x)-f(y)=xf(x)-yf(y)$ $\Rightarrow xf(y)=yf(x)$ $\Rightarrow f(x)=xf(1)$ let $f(1)=k$and $k\in R$ $\Rightarrow f(x)=kx$ it`s good but i am not safe

Writing the functional equation in the form $f(xx - yy) = xf(x) - yf(y)$, we see that all linear functionals from R to R would suffice, thus, since all functions of the form $f: \mathbb R \rightarrow \mathbb R, x \rightarrow kx$ are linear functionals, all such functions f would satisfy the functional equation. I do have a question, though: do there exist linear functionals from R to R that are not linear functions?

http://www.kalva.demon.co.uk/usa/usoln/usol024.html

mathisfun1 wrote: Writing the functional equation in the form $f(xx - yy) = xf(x) - yf(y)$, we see that all linear functionals from R to R would suffice, thus, since all functions of the form $f: \mathbb R \rightarrow \mathbb R, x \rightarrow kx$ are linear functionals, all such functions f would satisfy the functional equation. I do have a question, though: do there exist linear functionals from R to R that are not linear functions? Well you could have showed that simply by plugging in $f(x) = kx$. Since there are no restrictions on the problem (continuity, boundedness, differentiability, etc.) it's very hard to come up with an "argumentative" solution since functions can pretty much do whatever they want without restrictions and you can come up with some really goofy stuff (as you may have discovered in an Analysis course)

(i) x=0 f(x)=0 (ii) x>0 f(x)=x^(1/2)f(x^(1/2)) f(x)=x^(1/2+1/4+1/8+...+1/(2^n))f(x^(1/(2^n))) lim (n->infinity) f(x) = xf(1)=x f(x)=x (iii)x<0 x+y=0 y>0 0=f(0)=xf(x)-yf(y) xf(x)=-x*f(-x)=x^2 f(x)=x sorry for my poor English

EDIT: Oh, sorry, MellowMelon. I have a slightly different idea now; does this work?

Anyways, are we allowed to say that this is the well-known Cauchy function, so the function must be linear?

The QuattoMaster 6000 wrote: Now, since the function maps any real number to another real number, we see that $ f$ is a continuous function. This is not always true. Take the function that's outputs $ 1$ given a rational and $ 0$ given an irratinoal.

The QuattoMaster 6000 wrote: Anyways, are we allowed to say that this is the well-known Cauchy function, so the function must be linear? That isn't necessarily true - you need continuity at one point or monotonicity, otherwise there are funny non-continuous solutions. Don't know if you're allowed to say that if you can prove continuity or monotonicity.

hi all, I had a question regarding this problem. Once you establish that f(0)= 0 (our function goes through the origin), it's odd, and f(x^2)=x*f(x), could you proceed with the following reasoning: consider two points A(x,f(x)) and B(x^2, f(x^2)). If you form a right triangles with each point and the origin, dropping perpendiculars, the two triangles would be similar (because the ratio of the x-coordinates is x and the ratio of the y-coordinates is also x, or 1/x) so the two points must be collinear. Because they were chosen arbitrarily, wouldn't all the points in the function be collinear, and thus only linear functions satisfy our condition. I myself am somewhat iffy on this reasoning, but I would just like to see if one of you could show me where I blatantly go wrong.

You can't get from a real number to any other one through just squaring and taking square roots. Your logic shows that $ a^{2^n-1} f(a) = f(a^{2^n})$ for any $ a$ and integer $ n$, but that won't show for instance that $ 2f(1) = f(2)$. Note the only information you're using is $ f(x^2) = xf(x)$, and a nonlinear function that satisfies this is $ f(x) = 0$ for $ |x| \leq 1$, $ f(x) = x$ for $ |x| > 1$. (Don't forget you don't know whether your function is continuous.) On another note, a much easier way to do your argument is to rewrite $ f(x^2) = xf(x)$ as $ f(x^2)/x^2 = f(x)/x$ and let $ g(x) = f(x)/x$ for $ x$. Then $ g(x^2) = g(x)$, which is the most information you got.

Some harder problems a) \[ f(x^3 - y^3) = x^2f(x) - y^2f(y) \] b) \[ f(x^5 - y^5) = x^2f(x^3) - y^2f(y^3) \]

We are solving a f.e. more general than Allnames proposed. Find $ f: \mathbb R\rightarrow \mathbb R$ such that $ f(x^n - y^n) = x^{n - m}f(x^m) - y^{n - m}f(y^m)$, where $ n > m > 0$ are naturals. It is trivial to show Cauchy equation holds. So we have $ f(bx + c) = bf(x),\forall x$, where $ b$ is any integer. Also we have $ f(x^n) = x^{n - m}f(x^m)$. Using the identity, we have $ \sum_{i = 0}^{n - 1}\binom{n - 1}i(x + i)^n( - 1)^{n - 1 - i} = c_1x + c_2$, where $ c_1 = n!$ and $ c_2$ are constants. Let $ a_i = \binom{n - 1}i( - 1)^{n - 1 - i}$, we have $ c_1f(x) + f(c_2) = f(c_1x + c_2) = f(\sum_{i = 0}^{n - 1}a_i(x + i)^n) \\ = \sum_{i = 0}^{n - 1}a_if((x + i)^n) \\ = \sum_{i = 0}^{n - 1}a_i(x + i)^{n - m}f((x + i)^m) \\ = \sum_{i = 0}^{n - 1}\big(\sum_{j = 1}^m a_i(x + i)^{n - m}d_{ij}f(x^j)\big)$, where $ d_{ij} = \binom m{j}i^{m - j}$. Now we let $ x = k*e$ for integer $ k$ and fix $ e$. Using $ f(x^j) = k^jf(e^j)$, we can treat both sides of the above equality as a polynomial in $ k$. Since it holds for infinitely many $ k$, the coefficients of both sides have to be equal. Now we equating the coefficients of $ k$, we get $ c_1f(e) = n!f(e) \\ = \sum_{i = 0}^{n - 1}a_i\big(f(e)d_{i1}i^{n - m} + ef(1)d_{i0}i^{n - m - 1}(n - m)\big) \\ = mf(e)\sum_{i = 0}^{n - 1}\binom{n - 1}i( - 1)^{n - 1 - i}i^{n - 1} + (n - m)ef(1)\sum_{i = 0}^{n - 1}\binom{n - 1}i( - 1)^{n - 1 - i}i^{n - 1} \\ = f(e)(n - 1)!*m + f(1)(n - 1)!*(n - m)$, where we use identity again. Now we can easily get $ f(e) = ef(1)$. Since $ e$ is arbitrary, we have $ f(x) = f(1)x$. Q.E.D. It is easy to see we can solve the generalized case where, for $ r\ne\pm1,r\in\mathbb Q$, 1) $ f(x + y) = f(x) + f(y)$; 2) $ f(x^r) = x^{r - 1}f(x),\forall x>0$.

$f(x^{2}-y^{2})=xf(x)-yf(y)\Rightarrow f(x^{2})=xf(x)$ Setting $x=-x$ in this relation we get $f(x^{2})=xf(x)=-xf(-x)$ Hence $f$ is odd The initial equation reduces to $f(x^{2}-y^{2})= f(x^{2})+f(-y^{2})$ from which we can conlude that $f(a+b)=f(a)+f(b)$ for every $a\geq0$ and $b\leq0$ But since f is odd we have $f(a+b)=f(a)+f(b)$ for every real $a,b$ In $f(x^{2})=xf(x)$ we set $x=x+1$ and since $f$ satisfies Cauchy we get $f(x)=xf(1)$ Setting this value in the initial equation we conclude $ f(x)=ax$ for every real $a$

ClaimThe only solutions are $f(x) = cx$ , $c \in \mathbb{R}$ Let $P(x,y)$ denote the assertion. From, $P(x,0)$ and $P(0,x)$ we have: $f(x^2) = xf(x)$ and $f(-x^2) = -xf(x)$ \newline $\implies f$ is odd and f(0) = 0 From this we now have $f(x^2-y^2)+f(y^2) = f(x^2)$ \newline This implies that $f$ is an additive function. Now make the substitution $P(x+1,y)$, obtaining: \begin{align*} f((x+1)^2) &= (x+1)f()x+1 \\ \implies f(x^2) + 2f(x) + f(1) &= (x+1)f(x) + (x+1)f(1) \\ \implies f(x) = xf(1) \end{align*}From this the conclusion is clear

Note that $f$ is an odd function, as \[f(x^2-y^2)=xf(x)-yf(y)=-(yf(y)-xf(x))=-f(y^2-x^2).\] Substituting $y=0$, we find $f(x^2)=xf(x)$. Hence $f$ must be additive as \[f(x^2-y^2)=xf(x)-yf(y)=f(x^2)+f(-y^2).\] Now if we plug in $\left(\frac{t+1}{2}, \frac{t-1}{2}\right)$, using the fact that $f\left(\frac{t-1}{2}\right) = f\left(\frac{t+1}{2}\right)-f(1)$, we get \begin{align*} f(t) &= \frac{t+1}{2} \cdot f\left(\frac{t+1}{2}\right) - \frac{t-1}{2} \cdot f\left(\frac{t-1}{2}\right) \\ &= \left(\frac{t+1}{2} - \frac{t-1}{2}\right) f\left(\frac{t+1}{2}\right) +f(1) \cdot \frac{t-1}{2} \\ &= \frac 12 \left(f(t+1)-f(1)+t \cdot f(1)\right) \\ &= \frac 12 \left(f(t)+t \cdot f(1)\right) \end{align*} Rearranging, we find $f(t)=t \cdot f(1) \implies \boxed{f(x)=cx, \quad c \in \mathbb{R}}$, with clearly works. $\blacksquare$

Taking $y=0$ gives $f(x^2)=xf(x)$. Swapping $x$ with $y$ and setting $y=0$ gives $f(-x^2)=-xf(x)$, which implies $f(x^2)=-f(-x^2)$, and thus $f$ is odd. Therefore, $f(0)=0$. Since $f(y^2)=-f(-y^2)=yf(y)$, we may rewrite the given as \[f(x^2-y^2)=f(x^2)-f(y^2)\Longleftrightarrow f(x^2+y^2)=f(x^2)+f(y^2)\]therefore $f$ is additive. Plugging in $x=a+1$ to $f(x^2)=xf(x)$, so \[f(a^2)+2f(a) + f(a) = (a+1)f(a)+(a+1)f(1),\]thus $f(a)=af(1)$, and so $f$ is linear. This gives the solutions to be of the form of $kx$ for all $k$, which obviously work. $\square$

$f\equiv 0$ and $f \equiv kx$ work. $P(0,0) \implies f(0) = 0$. $P(x,0) \implies f(x^2) = xf(x)$ and $P(-x,0)\implies f(x^2) = -xf(-x)$. Thus we can combine these to get $f(-x) = -f(x)$ for all $x\neq 0$, that is $f$ is odd. Now, \begin{align*} &f(x^2 - y^2) = xf(x) - yf(y) = f(x^2) - f(y^2)\\ \implies &f(x^2 (-y^2)) = f(x^2) + f(-y^2)\\ \implies &f(a+b) = f(a) + f(b) \; \forall a\in\mathbb R^+, b\in\mathbb R^- .\end{align*} Now $f(x+y) = f(x) + f(y)$ is clearly true if either of them is $0$. Otherwise if $a\in\mathbb R^+$ and $b\in \mathbb R^-$, we then also have $f(a+b) = f(a) + f(b)$. Now first we show that $f(a+b) = f(a) + f(b)$ for all $a,b\in \mathbb R^+$. Then we just use the odd property of the function to derive that $f(a+b) = f(a) + f(b)$ for all $a,b\in \mathbb R^-$. We will then combine all these information to conclude that $f$ is Cauchy. So first let's prove $f(a+b) = f(a) + f(b)$ for all $a,b\in \mathbb R^+$. Pick some $c\in \mathbb R^+$ such that $c>b$. So, \begin{align*} f(a+b) &= f((a+c) + (b-c)) = f(a+c) + f(b-c)\\ &=f(a+c)+f(b+(-c))\\ &=f(a+c) + f(b) + f(-c)\\ &=[f(a+c) + f(-c)] + f(b)\\ &=f((a+c) + (-c)) + f(b)\\ &=f(a) + f(b) .\end{align*} Thus we now conclude that $f$ is indeed Cauchy! Now note that we have $f(x^2) = xf(x)$. So, $f((a+b)^2) = (a+b)f(a+b)\implies af(a) + f(2ab) + bf(b) = af(a) + af(b) + bf(a) + bf(b)$ from which we get $f(ab) + f(ab) = af(b) + bf(a)$. Now set $b=1$ to get $f(a) + f(a) = af(1) + f(a)\implies f(a) = af(1) = ka$ and we are done.

plug in $y=0$ to get $f(x^2)=xf(x)$, and $x=0$ to get $f(-y^2)=-yf(y)$ we can easily get that $f(0)=0$, and $f(a)=-f(-a)$ plug these into the original equation for $f(x^2-y^2)=f(x^2)-f(y^2)$ and $f(a-b)=f(a)-f(b)$ we then get that $f(x)$ is linear with $f(0)=0$, so $f(x)=kx$

Plug in $(x,0)$ to get that $f(x^2)=xf(x)$ So $f(x^2-y^2)=f(x^2)-f(y^2)$ Plug in $(-x,0)$ to get that $f(x^2)=-xf(-x)=xf(x)$, so $f(x)=-f(-x)$ and $f$ is odd So $f(x^2)-f(y^2)=f(x^2)+f(-y^2)=f(x^2-y^2)$, so $f$ is additive for a nonnegative and a nonpositive. Let $x,y=a,b$ where $a \le b$ $f(x^2-y^2)+f(2y^2)=f(x^2+y^2)$ because $x^2-y^2$ is nonpositive but $2y^2$ is nonnegative. $f(x^2+y^2)=f(2y^2)+f(x^2)-f(y^2)=f(2y^2)+f(-y^2)+f(x^2)=f(y^2)+f(x^2)$, so $f$ is additive for two nonnegatives. Because $f$ is even, $f$ is also additive for two nonpositives. So $f$ is additive. Because $f(x^2)=xf(x)$, $af(a)+f(2a)+f(1)=f(a^2+2a+1)=(a+1)f(a+1)=af(a+1)+f(a+1)=(a+1)f(a)+(a+1)f(1)$ $f(a)+af(1)=f(2a)$ $f(2a)-f(a)=af(1)$ $f(a)=af(1)$, so the answer is $\boxed{f(x)=kx}$

Solved with Orthogonal.. We claim the only functions are $\boxed{f(x)\equiv cx}$ for $c\in\mathbb{R}$. It is easy to check that these work. Let $P(x,y)$ denote the given assertion. $P(0,0)$ gives $f(0)=0$ and $P(x,0)$ gives $f(x^2)=xf(x)$. From $P(x,x)$ we get $f(x)=-f(-x)$. Thus $P(x,y)$ becomes $f(x^2-y^2)=f(x^2)+f(-y^2)$. It follows that \[ f(x+y)=f(x)+f(y)\ \ \ \ \ \ \ \forall(x,y)\in\mathbb{R}^+\times\mathbb{R}^-.\tag{*} \] Claim. We have $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. Proof. If $x$ and $y$ are both positive, $f(x+y)-f(x)=f(x+y)+f(-x)=f(y)$ so we are done. If exactly one of $x$ and $y$ are positive, we are done by $(*)$. If $x$ and $y$ are both negative, \[ f(x+y)=-f(-x-y)=-f(-x)-f(-y)=f(x)+f(y) \]by case $1$ so we are done. $\square$ Now \begin{align*} xf(x)+xf(1)+f(x)+f(1)&=(x+1)f(x+1)\\ &=f((x+1)^2)\\ &=f(x^2+2x+1)\\ &=xf(x)+2f(x)+f(1) \end{align*}so $f(x)=f(1)x$, as desired. $\square$

We claim that the solutions are $f(x)=cx$ for any constant $c$, and it's easy to see that this satisfies the equation. Plugging in $y=0$, we get $f(x^2)=xf(x)$, and plugging in $x=0$, we get $f(-y^2)=-yf(y)$. From this, we get $f(-x^2)=-xf(x)=-f(x^2)$, giving $-f(x)=f(-x)$. Now, using the first two equations we got by plugging in $x=0$ and then $y=0$, we get $f(x^2-y^2)=f(x^2)-f(y^2)$, which further becomes $f(x^2-y^2)=f(x^2)+f(-y^2)$ since $-f(x)=f(-x)$. This implies that if $x\geq 0, y\leq 0$ or $x\leq 0, y\geq 0$, we have $f(x)+f(y)=f(x+y)$. However, we can also get $f(x^2-y^2)+f(y^2)=f(x^2)$, and setting $x>y$ gives us that $f(x)+f(y)=f(x+y)$ also holds true when both $x,y\geq 0$. Finally, by negating everything and using $-f(x)=f(-x)$, this also holds true when both $x,y\leq 0$. Hence, $f$ satisfies Cauchy. Then, we plug in $x+1$ into $f(x^2)=xf(x)$ and use the fact that $f$ satisfies Cauchy to get \[f((x+1)^2)=f(x^2)+f(x)+f(x+1)=(x+1)f(x+1)=xf(x)+xf(1)+f(x)+f(1),\]and by using $f(x^2)=xf(x)$, we can cancel things to get $f(x+1)=f(1)(x+1)$, giving $f(x)=cx$ where $c=f(1)$.

We will show that the answer is $f(x) = cx$. These work since $(x^2 - y^2).c = x.cx - y.cy$. Now we need to show these are the only solutions. Let x = 0, we get $f(-y^2) = -yf(y)$. Let y = 0, we get $f(x^2) = xf(x)$. By these two we get that f is odd and also that $f(0) = 0$. So now using what we got we can write the starting equation as $f(x^2 - y^2) + f(y^2) = f(x^2)$. Using this and the fact that f is odd we get that f is additive. Now plugging in x = x + 1 in $f(x^2) = xf(x)$ and using that f is additive we get $f((x+1)^2 ) = (x+1)f(x+1)$ $\Rightarrow$ $f(x^2 + 2x + 1) = f(x^2) + 2f(x) + f(1) = (x+1)f(x) + (x+1)f(1)$ which after clearing things up gives us that $f(x) = f(1)x = cx$ $\Rightarrow$ there are no other solutions and we are ready.

We claim that the only solution is $\boxed{f(x) = cx}$ for some constant $c$. It is easy to verify that this satisfies the given condition. Now we will prove that it is the only solution. $\underline{P}(x, x) \implies \boxed{f(0) = 0}$ $\underline{P}(x, y) - \underline{P}(y, x) \implies f(y^2 - x^2) = - f(x^2 - y^2) \implies \boxed{f \text{ is odd}}$. $\underline{P}(x, 0) \implies \boxed{f(x^2) = x f(x)}$ $\implies f(x^2 - y^2) = f(x^2) - f(y^2)$ $\implies \boxed{f(x - y) = f(x) - f(y) \text{ for positive } x, y} \ldots \textcircled{1}$. Now choose some positive $x$. $\underline{P}(x - 1, x) \implies f(1 - 2x) = (x - 1)f(x - 1) - xf(x)$ $\overset{\textcircled{1}}{\implies} f(1) - f(2x) = (x - 1)(f(x) - f(1)) - xf(x)$ $\implies f(2x) - f(x) = xf(1)$ $\overset{\textcircled{1}}{\implies} f(x) = xf(1)$ So $f(x) = cx$ for positive $x$. Since $f$ is odd, this is also true for negative $x$. We are done.

From $f(x^2) = xf(x)$, and $f$ is odd, we can directly use Cauchy to finish

alexanderhamilton124 wrote: From $f(x^2) = xf(x)$, and $f$ is odd, we can directly use Cauchy to finish wait what?

I don't know if it's well known (somebody told me it was in BJV (?)), but look at the last property here

alexanderhamilton124 wrote: I don't know if it's well known (somebody told me it was in BJV (?)), but look at the last property here Here is another problem you can use this trick. The problem is from 2022 Kosovo TST P1 https://artofproblemsolving.com/community/c6h2797116p24625420