math154 wrote: Find all functions $f:\mathbb{R}^+\to\mathbb{R}^+$ such that whenever $a>b>c>d>0$ and $ad=bc$, \[f(a+d)+f(b-c)=f(a-d)+f(b+c).\] Calvin Deng. First, it is easy to see that when $f(x)$ is a positive constant satisifies the condition. We are going to prove that $f(x)=px^2+q$, where $p,q>0$ are the unique solution for the problem. Let $ad=k$, from $f(a+d)-f(a-d)=f(b+c)-f(b-c)$, we know that for all $a>\sqrt{k}$, we have $f(a+\frac{k}{a})-f(a-\frac{k}{a})=g(k)$, where $g(k)$ is another function when $k$ are positive real number. Let $a+\frac{k}{a}=x$, $a-\frac{k}{a}=y$, then $a=\frac{x+y}{2}$, $k=\frac{x^2-y^2}{4}$ Therefore for all $x>y$, we have $f(x)-f(y)=g(\frac{x^2-y^2}{4})$. Subsitute $x=\sqrt{y^2+4t}$ for some $t>0$ onto the equation. We have $f(\sqrt{y^2+4t})-f(y)=g(t)$, for every $y,t>0$. Assume there exists a $t$ such that $g(t)<0$, by induction we have $f(\sqrt{y^2+4nt})-f(y)=ng(t)$, for all positive integer $n$. Therefore when $n$ goes to infinity, $f(\sqrt{y^2+4nt})$ goes to negative. Contradiction! As a result, we have $g(t)\ge 0$, which means $f(x)$ is a non-decreasing function. At last, we use the subsitution $h(x)=f(\sqrt{x})$. We have $h(x+4t)-h(x)=g(t)$, for all $x,t>0$. By the Caushy functional equation and $f(x)$ is non-decreasing function, it implies that $h(x)$ is a linear function, let $h(x)$ be $px+q$, then $f(x)=px^2+q$, where $p,q>0$ After checking, $f(x)=px^2+q$, where $p\ge 0$, $q>0$ are the unique solution for the problem.

Your solution is very interesting but I think there is a little mistake here Aluminum wrote: At last, we use the subsitution $h(x)=f(\sqrt{x})$. We have $h(x+4t)-h(x)=g(t)$, for all $x,t>0$. By the Caushy functional equation and $f(x)$ is non-decreasing function, it implies that $h(x)$ is a linear function, let $h(x)$ be $px+q$, then $f(x)=px^2+q$, where $p,q>0$ After checking, $f(x)=px^2+q$, where $p\ge 0$, $q>0$ are the unique solution for the problem. You say that h is linear and you put the formula of an affine function. And I've not really understood how you used Cauchy... Can you ex plain me ?

apluscactus wrote: Your solution is very interesting but I think there is a little mistake here Aluminum wrote: At last, we use the subsitution $h(x)=f(\sqrt{x})$. We have $h(x+4t)-h(x)=g(t)$, for all $x,t>0$. By the Caushy functional equation and $f(x)$ is non-decreasing function, it implies that $h(x)$ is a linear function, let $h(x)$ be $px+q$, then $f(x)=px^2+q$, where $p,q>0$ After checking, $f(x)=px^2+q$, where $p\ge 0$, $q>0$ are the unique solution for the problem. You say that h is linear and you put the formula of an affine function. And I've not really understood how you used Cauchy... Can you ex plain me ? It is because I am a bit of lazy... When we have $h(x+4t)-h(x)=g(t)$ for all $x,t>0$ By interchanging $x$ and $4t$, we have $h(x+4t)-h(4t)=g(\frac{x}{4})$, which means for all $x,t>0$, $h(x)+g(\frac{x}{4})=h(4t)+g(t)$, and which implies $h(4t)+g(t)=c$, which is a constant. By subsituting it onto the original equation, we get $h(x+4t)-h(x)=c-h(4t)$. Indeed, it is the form of Caushy equation after subsitution of the $h(x)+c$. Therefore it is linear function and so we can let it be $h(x)=px+q$

Let $w$, $x$, $y$, $z$ be such that $w>y>z>x>0$ and $z+y>w+x$. Let $a=\frac{w+y}{2}$, $b=\frac{x+z}{2}$, $c=\frac{z-x}{2}$, $d=\frac{w-y}{2}$. It is not difficult to prove that $a>b>c>d>0$. Making this substitution into the given equation gives: $f(w)+f(x)=f(y)+f(z)$ for all $w>y>z>x>0$ such that $z+y>w+x$ and $w^2+x^2=y^2+z^2$. Let $f(p)=g(p^2)$ for all $p>0$. Then $g(w^2)+g(x^2)=g(y^2)+g(z^2)$ for all $w>y>z>x>0$ such that $z+y>w+x$ and $w^2+x^2=y^2+z^2$. We can discount $z+y>w+x$ as this is implied by $w>y>z>x>0$ and $w^2+x^2=y^2+z^2$. This implies that $g(w^2)+g(x^2)=g(y^2)+g(z^2)$ for all pairwise distinct $w, x, y, z>0$ such that $w^2+x^2=y^2+z^2$. Now make the substitution $a=\frac{w+y}{2}$, $b=\frac{w+z}{2}$, $c=\frac{w-z}{2}$, $d=\frac{y-w}{2}$, where $y>w>z>0$ and $2w^2=y^2+z^2$. Thus $a$, $b$, $c$, $d$ satisfy the required conditions, and this substitution is allowed. This gives $g(y^2)+g(z^2)=2g(w^2)$ for all $y>w>z>0$ such that $2w^2=y^2+z^2$. Summing up, we can conclude that $ g(w^2)+g(x^2)=g(y^2)+g(z^2)$ holds for all $w, x, y, z>0$ such that $w^2+x^2=y^2+z^2$. The solution is now completed by http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=487697 !

Answer: $f(x) = k_1 x^2 + k_2 $ for some $k_1, k_2 \ge 0$- and not both zero. Clearly these work. To see these are the only solutions, write this as \[ f(a+d) - f(a-d) = f(b+c) - f(b-c). \]Thus we can define a function $g : {\mathbb R}^+ \to {\mathbb R}$ such that \[ f(a+d) - f(a-d) = g(4ad) \]say (since the left-hand side depends only on $ad$). Thus for $x > y$ we have an identity \[ f(x) - f(y) = g(x^2-y^2). \]Now, by noting that \begin{align*} f(x) - f(y) &= g(x^2-y^2) \\ f(y) - f(z) &= g(y^2-z^2) \\ f(x) - f(z) &= g(x^2-z^2) \end{align*}we conclude that $g$ is additive. Consequently, from $f(x)-f(y) = g(x^2)-g(y^2)$ we actually have \[ g(x^2) - f(x) = k \]for some constant $k$. As $f \ge 0$, $g : {\mathbb R}^+ \to {\mathbb R}$ is bounded below, thus $g$ is linear. Finally we conclude $f$ has the form above.

v_Enhance wrote: Answer: To see these are the only solutions, write this as \[ f(a+d) - f(a-d) = f(b+c) - f(b-c). \]Thus we can define a function $g : {\mathbb R}^+ \to {\mathbb R}$ such that \[ f(a+d) - f(a-d) = g(4ad) \]say (since the left-hand side depends only on $ad$). Thus for $x > y$ we have an identity \[ f(x) - f(y) = g(x^2-y^2). \]Now, by noting that \begin{align*} f(x) - f(y) &= g(x^2-y^2) \\ f(y) - f(z) &= g(y^2-z^2) \\ f(x) - f(z) &= g(x^2-z^2) \end{align*}we conclude that $g$ is additive. How is g additive?

@above $$g(x^2-y^2)+g(y^2-z^2)=\bigg(f(x)-f(y)\bigg)+\bigg(f(y)-f(z)\bigg)=f(x)-f(z)=g(x^2-z^2)=g\Bigg(\Big(x^2-y^2\Big)+\Big(y^2-z^2\Big)\Bigg).$$

If $x<y$ and $f(x)>f(y)$, consider the sequence $a_n$ defined by $a_0=y+1$, and $(a_n-a_{n-1})(a_n+a_{n-1})=(y-x)(x+y)$. Then taking $(2a,2b,2c,2d)=(a_n+a_{n-1},y-x,x+y,a_n-a_{n-1})$ and applying the condition, we get that \[f(a_n)-f(a_{n-1})=f(y)-f(x)<0\]Therefore, $f(a_n)<0$ for sufficiently large $n$, which is impossible. Because $f$ is nondecreasing, it has at most countably many discontinuous points. I claim now that $f$ is continuous at every $x$. Let $y\ne x$ be a continuous point. Then for every $\varepsilon$ there exists a $\delta$ so that $(y-x\delta,y+x\delta)$, $(x-y\delta,x)$ and $(x,x+y\delta)$ are all continuous and \[f(y+\delta x)-f(y-\delta x)<\varepsilon\]Putting $(a,b,c,d)=(y,x,y\delta,x\delta)$ (or $(x,y,x\delta,y\delta)$ if $x$ is greater), \[f(x+\delta y)-f(x-\delta y)=f(y+\delta x)-f(y-\delta x)<\varepsilon\]so $f$ is continuous at every $x$. Now note that, for all $x$ and $y$, putting $(a,b,c,d)=(x,y,x\varepsilon,y\varepsilon)$, \[yf'(x)=y\lim_{\varepsilon\to 0} \frac{f(x+y\varepsilon)-f(x-y\varepsilon)}{2y\varepsilon}=x\lim_{\varepsilon\to 0} \frac{f(y+x\varepsilon)-f(y-x\varepsilon)}{2x\varepsilon}=xf'(y)\]Therefore, if they exist $\dfrac{f'(x)}{x}=\dfrac{f'(y)}{y}$. But the limits above cannot always be infinite, so the derivatives must exist. Now $f'$ is linear, so $f(x)=ax^2+c$ for some $a$ and $c$; obviously they are both nonnegative and not both 0.

I believe this is a new solution. Take $a=\frac{\left(\sqrt{2}+1\right)x^2-\left(\sqrt{2}-1\right)y^2}{2\sqrt{2}}, b=\frac{x^2+y^2+2\sqrt{2}xy}{2\sqrt{2}}, c=\frac{x^2+y^2-2\sqrt{2}xy}{2\sqrt{2}}, d=\frac{\left(\sqrt{2}-1\right)x^2-\left(\sqrt{2}+1\right)y^2}{2\sqrt{2}}$, where $x>\left(\sqrt{2}+1\right)y>0$, then we have $$f(x^2-y^2)+f(2xy)=2f\left(\frac{x^2+y^2}{\sqrt{2}}\right).$$ The system $x^2-y^2=u$ and $2xy=v$ has always solution for any $u>v>0$ so, for any $u>v>0$ we have, $$f(u)+f(v)=2f\left(\sqrt{\frac{u^2+v^2}{2}}\right).$$Lets take $f\left(\sqrt{t}\right)=g(t)$ then for any $u>v>0$ we have, $$g(u^2)+g(v^2)=2g\left(\frac{u^2+v^2}{2}\right).$$ Taking $r=u^2$ and $s=v^2$ so, for any $r>s>0$ we have, $$g(r)+g(s)=2g\left(\frac{r+s}{2}\right).$$Clearly last relation is true also for $s>r>0$ and it is clear also it is true for $r=s$ hence, for any positive real numbers $r, s$ we have, $$g(r)+g(s)=2g\left(\frac{r+s}{2}\right).$$Since $f$ is bounded from below it means also $g$ is bounded from below and since it is Jensen functional equation it means that $g(x)=kx+l$ for any positive real number $x$ and any positive constants $k,l$ so, $f(x)=kx^2+l$ for all positive real numbers $x$ and any positive constants $k,l$ which clearly satisfy the initial condition.

Rearrange to\[f(a+d) - f(a-d) = f(b+c) - f(b-c)\]which tells us that for any fixed $k$, across all $x, y$ multiplying to $k$, function $f(x+y) - f(x-y)$ is constant. Let $g(k) = f(x+y) - f(x-y)$ for all $xy = k$. If we let $x = m + n$ and $y = m - n$, then we get\[g(m^2 - n^2) = f(2m) - f(2n)\]for all $m, n$. Consequently, $g(n^2 - k^2) = f(2n) - f(2k)$ for all $n, k$, and adding yields\[g(m^2 - n^2) + g(n^2 - k^2) = f(2m) - f(2k) = g(m^2 - k^2)\]and since $m^2 - n^2, n^2 - k^2$ can take any real values, $g(a) + g(b) = g(a+b)$ across all reals. Thus,\[g(m^2 - n^2) = g(m^2) - g(n^2) = f(2m) - f(2n) \implies g(x^2) - f(2x) \text{ is constant for real $x > 0$.}\]Nicely enough, this bounds $g$ for us, as on positive domain, $g$ is bounded below by $f + c \geq c$ for some constant $c$. Since $g$ is both additive and bounded on some interval, hence is linear. Thus,\[g(x^2) - f(2x) = c \implies f(2x) = g(x^2) - c = c_1x^2 + c_2\]for some constants $c_1, c_2$. Upon plugging this back into the original assertion, we find that the restrictions on $c_1, c_2$ are that both are $\geq 0$ since the range of $f$ is positive, furthermore both cannot be $0$ at the same time. Our answer is all $f(x) = c_1x^2 + c_2$ for constants $c_1, c_2 \geq 0$ where at least one of them is $> 0$, and we are done. $\blacksquare$

The answer is $f(x) = kx^{2} + c$, for positive $k,c$ Consider $d = a-r, c=r-b$. Then, we have $f(2a-r) + f(2b-r) = 2f(r)$, if $a(a-r) = b(r-b)$. Solving for $a$ we have \[a^{2} - ra - b(r-b) = 0\Rightarrow a = \frac{r+\sqrt{r^{2}+4br-4b^{2}}}{2}\]Observe that $a > b$ since we assume $b < r$, and $a > r$. Next, observe that $d = a-r < c$, because \[-r + \sqrt{r^{2} + 4br-4b^{2}} < 2r - 2b \Leftrightarrow r^{2} + 4br - 4b^{2} < 9r^{2} - 12br + 4b^{2} \Leftrightarrow 8(r-b)^{2}\geq 0\]Therefore this value of $a$ satisfies $a(a-r) = b(b-r), a > b > c > d > 0$. Next, this means \[f(2a-r) + f(2b-r) = 2f(r)\Rightarrow f(\sqrt{r^{2} + 4rb - 4b^{2}}) + f(2b-r) = 2f(r)\]For any values of $x, y$, we can find some value of $b,r$ such that $x = \sqrt{r^{2} + 4rb - 4b^{2}}, y = 2b-r$ (we can take $2r^{2} = x^{2} + y^{2}$, then $b = \frac{y+r}{2}$). Since \[2r^{2} = (\sqrt{r^{2} + 4rb - 4b^{2}})^{2} + (2b-r)^{2}\]This means we now have the equation \[f(\sqrt{x}) + f(\sqrt{y}) = 2f\left(\sqrt{\frac{x^{2} + y^{2}}{2}}\right)\]Let $g(x) = f(\sqrt{x})$. We have $g(x) + g(y) = 2g(\frac{x+y}{2})$. This is Jensen's functional equation, and since $g$ is bounded, $g(x) = kx+c$ for positive $k, c$. Therefore, $f(x) = kx^{2} + c$.

The below solution is Isomorphic to post #10, but appears to a be a bit easier to read: The answer is $f(x) = cx^2 + d$ for any constants $c,d \in \mathbb R_{\ge 0}$ with $(c,d) \ne (0,0)$. These clearly work. Lemma: For any $p > q > 0$, we can choose $a > b > c > d > 0$ satisfying $ad = bc$ and $$a+d = \sqrt{p} ~,~ b-c = \sqrt{q} ~,~ a-d = b+c = \sqrt{\frac{p+q}{2}}$$ Proof: The existence of such $a,b,c,d$ without $ad = bc$ should be direct. Then to prove these $a,b,c,d$ also satisfy $ad = bc$ follows by noting $$4ad = (a+d)^2 - (a-d)^2 = \frac{p-q}{2} = (b+c)^2 - (b-c)^2 = 4bc \qquad{\square}$$ So now we have $$f(\sqrt{p}) + f(\sqrt{q}) = 2 f \left( \sqrt{\frac{p+q}{2}} \right) ~ \forall ~ p,q \in \mathbb R^+$$Consider the function $g : \mathbb R^+ \to \mathbb R^+$ defined by $g(x)= f(\sqrt{x})$. It suffices to show $g(x) \equiv cx + d$ for some constants $c,d$ (the bounds on $c,d$ follow easily). Now $(1)$ becomes $$g(p) + g(q) = 2 g \left( \frac{p+q}{2} \right) ~ \forall ~ p,q \in \mathbb R^+$$So $g$ satisfies the Jensen's F.E. , and since $g$ is $> 0$ everywhere, so we are done. $\blacksquare$

Let $g(x)=f(\sqrt x)$. Then \[g(a^2+d^2+2ad)+g(b^2+c^2-2bc)=g(a^2+d^2-2ad)+g(b^2+c^2+2bc).\]Now note that all this requires is $a^2+d^2>b^2+c^2>2bc$, so we can rewrite as \[g(x+z)+g(y-z)=g(x-z)+g(y+z).\]So let $h(z)=g(x+z)-g(x-z)$ whenever $x>z$. Since $g$ is over the positive reals, $h$ is always nonnegative. Let $S$ be the set of all rationals with denominator a power of $2$. Now let $g(1)=m$ and $g(2)=n$. Then $g(3)=2n-m$, $g(4)=3n-2m$, etc. Also, $g\left(\frac32\right)=\frac12 (m+n)$, $g\left(\frac12\right)=\frac12 (3m-n)$, etc. We can extend this to $S$, and therefore $g$ is linear over $S$. But now consider any real $r$. Since $S$ is dense over the reals, we must have some arbitrarily close $s,t\in S$ such that $s\le r\le t$. But since $g$ is nondecreasing, we must have $g(s)\le g(r)\le g(t)$. Since $g(s)$ and $g(t)$ can be arbitrarily close to each other, $g(r)$ must satisfy the linear pattern as well. Thus $g$ is linear, and $\boxed{f(x)=ax^2+b}$ for all nonnegative reals $a,b$ that are not both zero. Checking, all of these work. QED.

Note that this is equivalent to \[ f(w) - f(x) = f(y) - f(z) \]holds whenever $w^2 - x^2 = y^2 - z^2 > 0$. Substitute $g(x^2) = f(x)$ to get that \[ g(a + n) - g(a) \]is independent of $a$ for $n > 0$. We can substitute $h(x) = g(x+1) - g(1)$ such \[ h(a)+h(b)=(g(a+b+1)-g(b+1))+(g(b+1)-g(1))=h(a+b). \] Since $h$ is not dense as $g \ge 0$, it follows that $h$, and thus $g$, is linear. The solution set is thus $f(x) = ax^2 + b$ for $a, b \ge 0$ and not both are zero.

I am astounded that I missed this for an hour. Bruh. The problem effectively states that the function \[ g(ad) = f(a+d) - f(a-d) \]is constant for $a > d$. Substitute $x=a+d$ and $y=a-d$ to get \[ g\left (\frac{x^2-y^2}{4} \right) = f(x) - f(y) \]for all $x > y$. It can then be concluded that $g$ is additive over $\mathbb R^+$ which yields $g$ linear due to the boundedness of the codomain. Particularly $kx^2 - ky^2 = f(x) - f(y)$ for $x > y$. This finishes it; we get $f(x) = kx^2 + q$ for $k,q > 0$, but also just the constant functions.

The answer is $f(x) = \boxed{k_1x^2+k_2}$ for nonnegative reals $k_1,k_2$, such that both cannot be $0$. Plugging this in confirms it works. Rearrange the equation to get \[f(a+d)-f(a-d) = f(b+c)-f(b-c),\] and define $g(x)$ such that \[f(x)-f(y) = g(x^2-y^2),\] which indeed lines up with the $ad=bc$ condition. Now, notice that \begin{align*} g(x^2-y^2) + g(y^2-z^2) &= (f(x)-f(y))+(f(y)-f(z)) \\ &= f(x)-f(z) = g(x^2-z^2) \\ &= g((x^2-y^2)+(y^2-z^2)). \end{align*} Therefore, we conclude $g$ is additive. Moreover, we now have \[f(x)-f(y) = g(x^2)-g(y^2) \implies g(x^2)-f(x) = g(y^2)-f(y)= c\] for a constant $c$, and $f$ is bounded above $0$. Thus, $g$ is bounded above $c$, which along with its additive properties, implies that $g$ is linear. We can easily derive the desired solution set from this.

Let $a=Px$, $b=Py$, $c=\frac{P}{y}$, $d=\frac{P}{x}$ where $P>0$ and $x>y>1$. Then, $f(P(x+\frac1{x}))+f(P(y-\frac1{y}))=f(P(x-\frac1{x}))+f(P(y+\frac1{y}))$. This means that $f(P(x+\frac1x))-f(P(x-\frac1x))$ is constant as $x>1$ changes. Therefore, $f(P\sqrt{n^2+4})-f(Pn)$ is constant as $n>0$ changes. Let $g(x)=f(\sqrt{x})$. Then, $g(n^2P^2+4P^2)-g(n^2P^2)$ is constant as $n>0$ changes. Therefore, let $g(x+D)-g(x)=h(D)$ for all $x,D>0$. If $h(D)<0$ for some $D>0$, then $g(1+D(\lceil\frac{g(1)+69420}{-h(D)}\rceil))<0$, which is not cool. Therefore, $h$ is a function from $\mathbb{R}^+$ to $\mathbb{R}_{\ge 0}$. Since $h(a)+h(b)=h(a+b)$, Cauchy gives $h(x)=Hx$ for some $H\ge 0$. Since $\frac{g(x+D)-g(x)}{D}=H$ for all $x,D>0$, $g(x)=Hx+C$ for $H,C\ge 0$. Since $g(x)=f(\sqrt{x})$, $\boxed{f(x)=Hx^2+C}$ for $H,C\ge 0$. Plugging this back in gives $H(a^2+b^2+c^2+d^2)+2C$ on both sides, as desired.