is selected inside acute 
so that 
and 
.
is chosen on ray 
so that 
.
be the midpoint of 
.
is tangent to the circumcircle of triangle 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
and 
be positive integers. Let 
be the set of integer points 
with 
and 
.
rectangles is called happy if each point in 
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Let 
, 
, 
, 
, 
be distinct positive integers such that 
. Show that 
is a composite number.
, 
, 
, 
, 
be distinct positive integers such that 
. Show that 
is a composite number.
is prime, but it didn't get anywhere...
Now we get that if 
The first three cannot be true by size, and for the next, assuming for contradiction, we get that 
Since 
Without loss of generality assume 
so that 
or 
However, this implies 
and 
which fails.
or 
share any prime factors with each other then the result is obvious. Assume everything is relatively prime and take some 
so that 
or 
Then 
,
or 
.
we have an easy time, and considering only 
s
,
and 
is even. The only case I couldn't resolve was 
both even. I also noted that 2 is a quadratic residue for the 
but I don't expect that to be useful. Maybe 1 or 2 points? Looks like I missed the main factorization though.
is larger than each of 
,
.
so much trying to do number theory in ![$\mathbb Z[\zeta_8]$](http://latex.artofproblemsolving.com/b/6/9/b69cf52636eb54c7196931c6c8b7dde10678f232.png)
.
or something idk
and 
,
is also an odd prime. Thus 
is even, so either 
or 
.
and 
,
,
,
.
for even 
for odd 
?
,
such that 
also be a solution, in which case 
![\[(a^4+b^4)c^2d^2-(c^4+d^4)a^2b^2=(ac+bd)(ac-bd)(ad+bc)(ad-bc)=e^5(cd-ab)(cd+ab)\]](http://latex.artofproblemsolving.com/9/2/9/929abe475320b5093242c8e381f58405c1ee8fa4.png)
Let 
,
is a prime.
or 
,
.
are all distinct, 
,
.
.
.![\[p^2=(ac+bd)^2=a^2c^2+2abcd+b^2d^2>ad+bc.\]](http://latex.artofproblemsolving.com/c/1/2/c12d0338fcde51c82f67bbc83744a04a7f23939b.png)
,
and 
.
or 
.
.
.
and 
or 
and 
.
,
.
,
and 
,
,
.
.
,
.
and 
,
.
,
and 
.
and 
However, this means that 
,
,
.
.
.
.
or 
.
,
which is not true.
.
,
which is not true.
,
which is not true.
.
or any of the other three. Using this, note that
Then 
Similarly, 
.
or 
,
so the only possibility is for 
.
.
,
,
which is clearly impossible.
be a prime number. Then we have 
,
,
.
.
,
,
,
,
.
.
,
,
.
.
,
,
.
,
,
,
Notice 
Thus 
Now 
means 
and clearly 
so 
and similarly 
WLOG assume 
then from 
we have 
so 
and 
contradiction.
![\[e^5 = a^4+b^4 \le a^5+b^5 < (ac+bd)^5 = p^5 \implies p>e.\]](http://latex.artofproblemsolving.com/6/4/2/642435f197980ea752ebf4637fa667547ce1e5eb.png)
and we similarly have 
.![\[a^2+d^2 \ge p\]](http://latex.artofproblemsolving.com/f/5/7/f575eb775d9024678fe7dd37f1aef85a7dfb1c17.png)
![\[b^2+c^2 \ge p.\]](http://latex.artofproblemsolving.com/3/b/7/3b712a367a1f0740733ac9885e0c4504f1f0d8ad.png)
,
,
,
,
.
,
.
,
.
,
.
,
.
,
,
,
,
,
,
as well, meaning 
,
doesn't divide 
.
.
and analogously 
.
and 
.
are distinct we must have either 
or 
.
Prove that 
is composite.
Clearly 
,
due to size reasons. Thus we have reduced to,
If 
,
or 
as 
and 
.
a contradiction.
which fails due to size reasons once more.
be a prime.![\[p | e^5(a-d)(a+d)\left(a^2+d^2\right).\]](http://latex.artofproblemsolving.com/8/5/0/850883379d079efdd628a8f0d1bcf6d0581c0b43.png)
.
,
hence we have 
,
implying 
which is a contradiction.
condition artificially stuff worked out.
(1)
.
by rearrangement, thus
(2)
,
,
