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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Consecutive squares are floors
ICE_CNME_4   11
N 12 minutes ago by ICE_CNME_4

Determine how many positive integers \( n \) have the property that both
\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]are consecutive perfect squares.
11 replies
ICE_CNME_4
Yesterday at 1:50 PM
ICE_CNME_4
12 minutes ago
Finding all possible $n$ on a strange division condition!!
MathLuis   10
N 35 minutes ago by justaguy_69
Source: Bolivian Cono Sur Pre-TST 2021 P1
Find the sum of all positive integers $n$ such that
$$\frac{n+11}{\sqrt{n-1}}$$is an integer.
10 replies
MathLuis
Nov 12, 2021
justaguy_69
35 minutes ago
IMO 2012 P5
mathmdmb   123
N 44 minutes ago by SimplisticFormulas
Source: IMO 2012 P5
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.

Show that $MK=ML$.

Proposed by Josef Tkadlec, Czech Republic
123 replies
mathmdmb
Jul 11, 2012
SimplisticFormulas
44 minutes ago
Fixed line
TheUltimate123   14
N an hour ago by amirhsz
Source: ELMO Shortlist 2023 G4
Let \(D\) be a point on segment \(PQ\). Let \(\omega\) be a fixed circle passing through \(D\), and let \(A\) be a variable point on \(\omega\). Let \(X\) be the intersection of the tangent to the circumcircle of \(\triangle ADP\) at \(P\) and the tangent to the circumcircle of \(\triangle ADQ\) at \(Q\). Show that as \(A\) varies, \(X\) lies on a fixed line.

Proposed by Elliott Liu and Anthony Wang
14 replies
TheUltimate123
Jun 29, 2023
amirhsz
an hour ago
No more topics!
IMO 2008, Question 4
orl   119
N May 5, 2025 by lpieleanu
Source: IMO Shortlist 2008, A1
Find all functions $ f: (0, \infty) \mapsto (0, \infty)$ (so $ f$ is a function from the positive real numbers) such that
\[ \frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}
\]
for all positive real numbers $ w,x,y,z,$ satisfying $ wx = yz.$


Author: Hojoo Lee, South Korea
119 replies
orl
Jul 17, 2008
lpieleanu
May 5, 2025
IMO 2008, Question 4
G H J
Source: IMO Shortlist 2008, A1
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ryanbear
1056 posts
#110 • 2 Y
Y by Spiritpalm, Amir Hossein
Plug in $y=w, z=x$
$f(x)^2+f(y)^2=f(x^2)+f(y^2)$
When $x=y$, get that $f(x)^2=f(x^2)$
So plugging in $x=1$, get that $f(1)=1$
If $w=1$, then $x=yz$
$\frac{1+f(yz)^2}{f(y^2)+f(z^2)}=\frac{1+y^2z^2}{y^2+z^2}$
If $y=z=\sqrt{x}$, then $\frac{1+f(x)^2}{2f(x)}=\frac{1+x^2}{2x}$
$2x+2xf(x)^2=2f(x)+2x^2f(x)$
$2xf(x)^2+f(x)(-2x^2-2)+2x=0$
$f(x)=x$ or $\frac{1}{x}$ by the quadratic formula
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eibc
600 posts
#111 • 1 Y
Y by Amir Hossein
The answers are $f(x) = x$ and $f(x) = \tfrac{1}{x}$, both of which aren't too hard to verify. We now show these are the only answers.

Let $P(w, x, y, z)$ denote the given assertion. From $P(1, 1, 1, 1)$, we have $f(1) = 1$. Then, from $P(1, x, \sqrt{x}, \sqrt{x})$, we have $\tfrac{1 + f(x)^2}{2f(x)} = \tfrac{1 + x^2}{2x}.$ This rearranges to
$$xf(x)^2 - f(x) - x^2f(x) + x = 0 \iff (xf(x) - 1)(f(x) - x) = 0.$$Thus, for all $x$, either $f(x) = \tfrac{1}{x}$ or $f(x) = x$. Now, suppose for the sake of contradiction there exist $a, b > 0$ and not equal to $1$ such that $f(a) = \tfrac{1}{a}$ and $f(b) = b$. Then from $P(a, b, 1, ab)$, we see that
$$\frac{\tfrac{1}{a^2} + b^2}{1 + f(a^2b^2)} = \frac{a^2 + b^2}{1 + a^2b^2}.$$Taking cases on whether $f(a^2b^2) = \tfrac{1}{a^2b^2}$ or $f(a^2b^2) = a^2b^2,$ we see that:
  • If $f(a^2b^2) = \tfrac{1}{a^2b^2}$, then $\tfrac{1}{a^2} + b^2 + b^2 + a^2b^2 = a^2 + b^2 + \tfrac{1}{a^2} + \tfrac{1}{b^2}$, or $(a^2b^2 + 1)(b^4 - 1) = 0$, which implies that $b = 1$, contradiction
  • If $f(a^2b^2) = a^2b^2$, then $\tfrac{1}{a^2} + b^2 + b^2 + a^2b^4 = a^2+ b^2 + a^2b^4 + a^4b^2$, or $(a^2b^2 + 1)(a^4 - 1) = 0$, so $a = 1$, contradiction
So, either $f(x) =x$ for all $x$, or $f(x) = \tfrac{1}{x}$ for all $x$, as desired.
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peppapig_
280 posts
#112 • 2 Y
Y by Amir Hossein, Jndd
I claim that the only solutions are $f(x)=x \forall x\in \mathbb{R}_{>0}$ and $f(x)=\frac{1}{x} \forall x\in \mathbb{R}_{>0}$.

(1) First, if we plug in $(c,c,c,c)$, we get that
\[\frac{2f(c)^2}{2f(c^2)}=1 \implies f(c)^2=f(c^2),\]since $f$ is over the positive reals. Additionally, we can also plug in $c=1$ to get that
\[f(1)^2=f(1) \implies f(1)=1,\]again since $f$ is over the positive reals.

(2) Now, if we plug in $(\sqrt{c},\sqrt{c},1,c)$, we get that
\[\frac{f(\sqrt{c})^2+f(\sqrt{c})^2}{f(1)+f(c^2)}=\frac{c+c}{1+c^2} \implies \frac{2f(c)^2}{f(1)+f(c)^2}=\frac{2c}{1+c^2},\]since we found by (1) that $f(x)^2=f(x^2)$. Since $c$ is fixed, letting $f(c)=k$, we can solve the equation
\[\frac{k}{1+k^2}=\frac{c}{1+c^2},\]\[\iff k(1+c^2)=c(1+k^2),\]\[\iff ck^2-(1+c^2)k+c=0,\]\[\iff (ck-1)(k-c)=0.\]This means that $k=c$ or $\frac{1}{c}$, which implies that $\forall x\in \mathbb{R}_{>0}$, $f(x)$ is either $x$ or $\frac{1}{x}$.

Now for the pointwise trap! I claim that either $f(x)=x \forall x\in \mathbb{R}_{>0}$ or $f(x)=\frac{1}{x} \forall x\in \mathbb{R}_{>0}$. FTSOC, assume that $\exists a$, $b\neq 1$ such that $f(a)=a$ and $f(b)=\frac{1}{b}$. Plugging in $(\sqrt{ab},\sqrt{ab},a,b)$ then gives us that
\[\frac{f(ab)+f(ab)}{f(a)^2+f(b)^2}=\frac{ab+ab}{a^2+b^2} \implies \frac{f(ab)}{a^2+\frac{1}{b^2}}=\frac{ab}{a^2+b^2}, \]\[\iff f(ab)=\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right).\]By (2), we know that $\forall x\in \mathbb{R}_{>0}$, $f(x)$ is either $x$ or $\frac{1}{x}$, which means that $f(ab)=ab$ or $\frac{1}{ab}$.

If $f(ab)=ab$, then we have that
\[\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right)=ab,\]\[\iff \frac{a^2b^2+1}{b^2(a^2+b^2)}=1,\]\[\iff a^2b^2+1=b^2(a^2+b^2) \implies a^2b^2+1=a^2b^2+b^4,\]\[\iff b^4=1,\]\[\implies b=1,\]since $f$ is over the positive reals. However, this is a contradiction to our requirement that $a$, $b\neq 1$. Therefore, we get that $f(ab)=\frac{1}{ab}$.

Now, if $f(ab)=\frac{1}{ab}$, we get that
\[\left(\frac{ab}{a^2+b^2}\right)\left(\frac{a^2b^2+1}{b^2}\right)=\frac{1}{ab},\]\[\iff ab(ab)(a^2b^2+1)=b^2(a^2+b^2),\]\[\iff a^4b^4+a^2b^2=a^2b^2+b^4,\]\[\iff a^4b^4=b^4,\]\[\implies a^4=1,\]\[\implies a=1,\]since $f$ is over the positive reals. This is again a contradiction to our requirement that $a$, $b\neq 1$, meaning that there cannot exist $a$, $b\neq 1$ such that $f(a)=a$ and $f(b)=\frac{1}{b}$. Therefore our only solutions are $f(x)=x \forall x\in \mathbb{R}_{>0}$ and $f(x)=\frac{1}{x} \forall x\in \mathbb{R}_{>0}$, which can both be checked to work through plugging and chugging, finishing the problem.
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KevinYang2.71
428 posts
#113 • 3 Y
Y by Amir Hossein, megarnie, deduck
Solved with blueberryfaygo_55.

We claim either $\boxed{f(x)\equiv x}$ or $\boxed{f(x)\equiv\frac{1}{x}}$. It is easy to check that these work.

Let $P(w,x,y,z)$ be the given assertion. Then $P(1,1,1,1)$ gives $f(1)=1$. $P(x,x,x,x)$ gives $f(x)^2=f(x^2)$. It follows that
\[
\frac{f(w^2)+f(x^2)}{f(y^2)+f(z^2)}=\frac{w^2+x^2}{y^2+z^2}
\]whenever $wx=yz$. Letting $(a,b,c,d):=(w^2,x^2,y^2,z^2)$, we have
\[
\frac{f(a)+f(b)}{f(c)+f(d)}=\frac{a+b}{c+d}\tag{*}
\]whenever $ab=cd$. Plugging in $(a,b,c,d)=(x,x,x^2,1)$ gives
\[
\frac{2f(x)}{f(x^2)+1}=\frac{2x}{x^2+1}
\]so
\[
(x^2+1)f(x)=xf(x)^2+x.
\]Solving this quadratic yields $f(x)\in\left\{x,\frac{1}{x}\right\}$.

Assume for the sake of contradiction there exists $m,n\in\mathbb{Z}^+\setminus\{1\}$ such that $f(m)=m$ and $f(n)=\frac{1}{n}$. Plugging $(a,b,c,d)=(m,n,mn,1)$ into $(*)$ gives
\[
\frac{m+\frac{1}{n}}{f(mn)+1}=\frac{m+n}{mn+1}.
\]Note that $f(mn)\in\left\{mn,\frac{1}{mn}\right\}$. If $f(mn)=mn$, then
\[
\frac{m+\frac{1}{n}}{mn+1}=\frac{m+n}{mn+1}
\]so $n=1$, a contradiction. If $f(mn)=\frac{1}{mn}$, then
\[
mn\cdot\frac{m+\frac{1}{n}}{1+mn}=\frac{m+n}{mn+1}
\]so $m=1$, a contradiction. The conclusion follows. $\square$
This post has been edited 1 time. Last edited by KevinYang2.71, Apr 26, 2024, 6:49 AM
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Mr.Sharkman
501 posts
#114
Y by
Taking $w=x=y=z=1$ gives us that
$f(1) = f(1)^{2},$
so $f(1) = 1.$ Also, $w = x = y = z = n$ for some real number $n$ gives us that $f(x^{2}) = f(x)^{2}.$
Now, setting $w = n^{2}, x = 1, y = n, z = n,$ gives us that
$$f(n) = n, \frac{1}{n}$$after solving. Now, examining all of the cases, we find that a pointwise trap cannot occur, so we are done.
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Jndd
1417 posts
#115
Y by
Our answers are $f(x)=x$ and $f(x)=\frac1x$, and it is easy to check that these satisfy our equation.

Plugging in $w=x=y=z$, we get $f(x)^2=f(x^2)$. Therefore, our equation becomes $\frac{f(w^2)+f(x^2)}{f(y^2)+f(z^2)}=\frac{w^2+x^2}{y^2+z^2}$, and since this is over the positive reals, we can replace $w^2$, $x^2$, $y^2$, and $z^2$ with $w$, $x$, $y$, and $z$ respectively, giving us \[\frac{f(w)+f(x)}{f(y)+f(z)}=\frac{w+x}{y+z}.\]Note that this substitution works with the condition $wx=yz$ because previously, this implied $w^2x^2=y^2z^2$, meaning replacing the variables still makes our equation hold true for $wx=yz$.

Plugging in $w=x=y=z=1$ into the original equation, we get $\frac{2f(1)^2}{2f(1)}=1$, giving $f(1)=1$. Now, plugging in $1$, $x^2$, $x$, and $x$ for $w$, $x$, $y$, and $z$ respectively, we get \[\frac{f(1)+f(x^2)}{2f(x)}=\frac{1+f(x^2)}{2f(x)}=\frac{1+x^2}{2x}.\]Clearing the denominators and factoring this, we get \[2(f(x)-x)(xf(x)-1)=0,\]which gives $f(x)=x$ and $f(x)=\frac1x$.

We will show that we cannot have $f(a)=a$ while $f(b)=\frac{1}{b}$ at the same time. Suppose $a,b\neq 1$, because otherwise we would have $a=\frac1a$ or $b=\frac1b$. Then, plugging in $a$, $b$, $1$, and $ab$ as $w$, $x$, $y$, and $z$, respectively, we get \[\frac{a+\frac1b}{1+f(ab)}=\frac{a+b}{1+ab},\]and by expanding and simplifying, it is easy to check that this equation does not hold for $f(ab)=ab$ or $f(ab)=\frac{1}{ab}$, so we are done.
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ezpotd
1286 posts
#116
Y by
The answers are $x, \frac 1x$, both of which clearly work.

Set $w = x = y = z = 1$, we get $f(1) = 1$. Now set $y = z = 1$, we get $f(w)^2+ f(\frac 1w)^2 = w^2 + \frac{1}{w^2}$, set $w = x = 1$ and we get $f(y^2) + f(\frac{1}{y^2}) = y^2 + \frac{1}{y^2}$. Thus we have $f(k)^2  + f(\frac 1k)^2 = k^2 + \frac{1}{k^2}$, as well as $f(k) + f(\frac 1k) = k + \frac 1k$, squaring the latter equation and subtracting from the first yields $f(k)f(\frac 1k) = 1$. Now using this with $f(k) + f(\frac 1k) = k + \frac 1k$ and solving the resulting quadratic gives $f(k) = k$ OR $f(k) = \frac 1k$ for each individual $k$. It remains to resolve the pointwise trap. We show that if there exist values such that $x \neq 1, f(x) = x$ there exist no values with $x \neq 1, f(x) = \frac 1x$, thus we must have either $f(x) = x$ or $f(x)  = \frac 1x$ for all $x$. Take $y = z, y \neq 1$, such that $f(y^2) = y^2$, this gives $f(w)^2 + f(x)^2 = w^2 + x^2$. We show that there exists no value of $x \neq 1$ satisfying $f(x) = \frac 1x$. Assume FTSOC this exists, then if $f(w) = w$, we are forced to have $x^4 = 1$, giving $x =1$, otherwise if $f(w) = \frac 1w$ we have $wx = 1$, contradiction since $wx = yz = y^2 \neq 1$.
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pie854
243 posts
#117
Y by
Here's a sketch:

Set $a=b=c=d=1$, the equation implies $f(1)=1$. Set $c=a$ and $d=b=1$ then we get $f(a^2)=f(a)^2$. Set $b=a$, $c=1$ and $d=a^2$ then after a bit of simplification we get $f(a)\in \{1/a,a\}$. Suppose $f(a)=a$ and $f(b)=1/b$ for some $a,b\neq 1$, setting $c=1$, $d=ab$ will give a contradiction (remember $f(ab)=ab$ or $f(ab)=1/(ab)$). Clearly $f(x)=x$ for all $x$ and $f(x)=1/x$ for all $x$ are solutions to the FE.
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bjump
1032 posts
#118
Y by
Solution
This post has been edited 1 time. Last edited by bjump, Oct 28, 2024, 7:31 PM
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Siddharthmaybe
117 posts
#119
Y by
P(t,t,t,t) gives f(t^2) = f(t)^2 or f(1) = 1 [I proved it in a lil non straight manner but Ic that this works too]
P(sqrt(t),sqrt(t), t , 1) gives:-
2f(t)/f(t)^2 + 1 = 2t/t^2+1
or tf(t)^2 - (t^2 + 1)f(t) + t = 0
Solving it as a quadratic (this is possible because t has a degree of freedom at all points) we get:-
FINAL ANSWER: f(t)= t, 1/t for all t in R+
Do Pointwise trap obviously otherwise ur stupid
both of which can be easily verified
This post has been edited 1 time. Last edited by Siddharthmaybe, Apr 3, 2025, 7:37 PM
Reason: ptwise
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Maximilian113
575 posts
#120
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Let $P(w, x, y, z)$ denote the assertion.

Then $P(x, x, x, x) \implies f(x^2)=f(x)^2 \implies f(1)=1.$ Also, the assertion then becomes $$\frac{f(w)+f(x)}{w+x} = \frac{f(y)+f(z)}{y+z}.$$Now, note that $$P(x, x, 1, x^2) \implies \frac{f(x)}{x} = \frac{1+f(x^2)}{1+x^2} = \frac{1+f(x)^2}{1+x^2}.$$Therefore, $$0=xf(x)^2-(1+x^2)f(x)+x \implies \left(f(x)-x \right)\left(f(x)-\frac{1}{x}\right)=0,$$so $f(x)$ is $x$ or $\frac{1}{x}$ for each $x \in (0, \infty).$ If there does not exist a $k$ with $f(k)=k,$ then $f(x)=\frac{1}{x}$ for all $x,$ which is a valid solution. Otherwise, for $k, a, b \neq 1$ in our domain with $ab=k^2$ and $f(k)=k,$ note that $$P(k, k, a, b) \implies f(a)+f(b)=a+b.$$If $f(a)=a,$ then $f(b)=b,$ and vice versa. If $f(a)=\frac{1}{a}, f(b)=\frac{1}{b},$ we have that $\frac{1}{a}+\frac{1}{b}=a+b \implies ab=1 \implies k=1,$ a contradiction. Therefore $f(x)=x$ for all $x,$ which is a valid solution.

Hence, we have determined that the only solutions to the functional equation are $f(x)=x, f(x)=\frac{1}{x}.$ It is easy to verify that they work.
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blueprimes
356 posts
#122
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imma crash out if i have to bash another pointwise trap

We claim the solutions are $f(w) \equiv w, \dfrac{1}{w}$ which can easily be checked to work. Now we prove these are the only ones, setting $x = z = 1$ we obtain $f(w)^2 + f(1)^2 = f(w^2) + f(1) \implies f(w)^2 = f(w^2) + c$ for some constant $c$. Substituting back into the given equation and replacing $w^2 \mapsto w$ and vice versa we obtain
\[\dfrac{f(w) + f(x) + 2c}{f(y) + f(z)} = \dfrac{w + x}{y + z} \]for all $wx = yz$. Let the above assertion be $P(w, x, y, z)$. Now $P(w, w, w, w)$ gives $c = 0$ then $f(1) - f(1)^2 = 0 \implies f(1) = 1$ and $f(w^2) = f(w)^2$ for all $w$. Now $P(w^2, 1, w, w)$ yields
\[ \dfrac{f(w^2) + f(1)}{2 f(w)} = \dfrac{f(w)^2 + 1}{2 f(w)} = \dfrac{w^2 + 1}{2w} \implies f(w) \in \left \{w, \dfrac{1}{w} \right \} \forall w. \]From here we solve the pointwise trap, for the sake of contradiction assume $a, b \ne 1$ exist where $f(a) = a, f(b) = \dfrac{1}{b}$. Then $P(a^2, b^2, ab, ab)$ yields
\[\dfrac{f(a^2) + f(b^2)}{2 f(ab)} = \dfrac{a^2 + b^2}{2ab} \implies a(a^2b^2 + 1) = b(a^2 + b^2) f(ab) \]from $f(a^2) = a^2$ and $f(b^2) = \dfrac{1}{b^2}$. If $f(ab) = ab$, the above relation simplifies to
\[a(a^2b^2 + 1) = ab^2(a^2 + b^2) \implies a^3b^2 + a = a^3b^2 + ab^4 \implies b^4 = 1 \implies b = 1\]which fails. On the other hand, if $f(ab) = \dfrac{1}{ab}$ we obtain
\[a^2b(a^2b^2 + 1) = b(a^2 + b^2) \implies a^4b^2 + a^2 = a^2 + b^2 \implies a^4 = 1 \implies a = 1 \]which also fails. Contradiction, so $f(w) \equiv w, \dfrac{1}{w}$ are our only solutions as wanted.
This post has been edited 1 time. Last edited by blueprimes, Feb 27, 2025, 2:36 AM
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Marcus_Zhang
980 posts
#123
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FE with quadratics??
This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 15, 2025, 1:31 AM
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Ilikeminecraft
658 posts
#124
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I claim the answer is $f(x) = x$ or $f(x) = \frac1x.$

First, let $w = x = y = z = 1.$ We see that $f(1)^2= f(1),$ and thus $f(1) = 1$ because the codomain is positive reals. Next, let $w = x = 1, yz = 1.$ Hence, we have that $f(y^2) + f\left(\frac1{y^2}\right) = y^2 + \frac1{y^2}.$ However, since our function is over the positive reals, we can assume that $f(x) + f\left(\frac1x\right) = x  +\frac1x.$ Now, we plug in $wx = 1, y = z = 1.$ Hence, $f(x)^2 + f\left(\frac1x\right)^2 = x^2 + \frac1{x^2}.$ By squaring the old equation and subtracting these two, we see that $f(x)f\left(\frac1x\right)= 1.$ Now, by solving this with the other equation, we see that $f(x) = x, \frac1x.$ To address pointwise, we can just trivially do casework to show that it can only be one.
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lpieleanu
3002 posts
#125
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Solution
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