IMO 2012 Problem 4
by liberator, Oct 5, 2018, 8:45 PM
This was a pretty tricky functional equation, especially for a P4...
Problem: Find all functions
such that, for all integers
that satisfy
, the following equality holds:
(Here
denotes the set of integers.)
Proposed by Liam Baker, South Africa
My solution
Problem: Find all functions



![\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]](http://latex.artofproblemsolving.com/a/9/e/a9edf1a3207d29facc970b6ee288d4fabdd43a4b.png)

Proposed by Liam Baker, South Africa
My solution
We claim that the only solutions are
,
and two pathological ones:
for any
.
First,
implies that
. Now
implies
, so we can rearrange the original equation to get
Let
denote the above equation. Then
, so
for each
.
Suppose
for some
; then
, so
is periodic with period
.
If
, we recover the solution
.
Hence suppose that
. If
, we get the solution
. Thus assume that
.
![[asy]
unitsize(0.6cm);
real e = 1;
MP("k",origin,E);
D((e,0)--(2,0));
MP("4k",(2,0),E);
D((4,-1)--(2+e,0)--(4,1));
MP("k",(4,1),E);
MP("9k",(4,-1),E);
D((6,0)--(4+e,1)--(6,2));
D((6,0)--(4+e,-1)--(6,-2));
DPA((6,0.3)--(7,-0.3)^^(6,-0.3)--(7,0.3),red);
MP("0",(6,2),E);
MP("\color{red}4k",(6,0),E);
MP("16k",(6,-2),E);
MP("f(1)",(0,3),E);
MP("f(2)",(2,3),E);
MP("f(3)",(4,3),E);
MP("f(4)",(6,3),E);
[/asy]](//latex.artofproblemsolving.com/4/c/b/4cbcba452f921e2ef6cad5f6377172915491d4fd.png)
By the tree diagram above, we have two cases:
and
. In the first case, we recover solution
, so assume that
.
We show by induction that
is the only solution in this case. Suppose that
for
. Then




First,




![$$\big[f(a)+f(b)-f(a+b)\big]^2=4f(a)f(b).$$](http://latex.artofproblemsolving.com/9/6/2/9628037cf5e447b568f9c5eee86d2de4f04c1147.png)

![$P(a,a)\implies f(2a)\big[f(2a)-4f(a)\big]=0$](http://latex.artofproblemsolving.com/e/6/0/e600d7841028e324f93ed64811d153132dfb8ba2.png)


Suppose





If


Hence suppose that




![[asy]
unitsize(0.6cm);
real e = 1;
MP("k",origin,E);
D((e,0)--(2,0));
MP("4k",(2,0),E);
D((4,-1)--(2+e,0)--(4,1));
MP("k",(4,1),E);
MP("9k",(4,-1),E);
D((6,0)--(4+e,1)--(6,2));
D((6,0)--(4+e,-1)--(6,-2));
DPA((6,0.3)--(7,-0.3)^^(6,-0.3)--(7,0.3),red);
MP("0",(6,2),E);
MP("\color{red}4k",(6,0),E);
MP("16k",(6,-2),E);
MP("f(1)",(0,3),E);
MP("f(2)",(2,3),E);
MP("f(3)",(4,3),E);
MP("f(4)",(6,3),E);
[/asy]](http://latex.artofproblemsolving.com/4/c/b/4cbcba452f921e2ef6cad5f6377172915491d4fd.png)
By the tree diagram above, we have two cases:




We show by induction that



- Comparing
and
implies
.
- Comparing
and
implies
.
- Comparing
and
implies
.
- Comparing
and
implies
.