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be three non-negative real numbers satisfying 
such that ![\[f(x+f(x)+2f(y))+f(2f(x)-y)=4x+f(y)\]](http://latex.artofproblemsolving.com/4/4/2/4423651bb829b92debd65a23f6cee3d4d4b8530f.png)
holds for all reals 
and 
.
, such that: 
For all 
.
are 
, 
, 
, 
, and 
. For 
, we take the following cases:
, take 
.
, take 
.
, let 
be the smaller solution in 
to 
, which exists since the discriminant is 
and is a quadratic residue mod 
. It is not hard to check that 
, so we can take 
.
for which![\[\left\{2^{-1}, 3^{-1}, \ldots, \left(\frac{p-1}{2}\right)^{-1}\right\} = \left\{-2, -3, \ldots, -\frac{p-1}{2}\right\} \pmod{p}.\]](http://latex.artofproblemsolving.com/e/a/b/eab1eb56436a33d87c2c5f77c799a7ee27dab26d.png)
which are easily checkable. Now we prove that any other prime is not mundane. Note that 
is clearly not mundane because of 
. Let 
.
then since , we can set 
and 
.
then since , we can set and 
.
then since , we can set 
and 
.
then since , we can set 
and 
, which works because 
and 
.
,
,
,
,
.
.
and 
less than 
such that 
is a positive integer. Find, with proof, all prime numbers that are not mundane.
check by hand that only 
and 
,![\[(a,b)=\begin{cases}
\left(\frac{2p-2}{5},\frac{p-5}{2}\right) & p \equiv 1 \pmod{5} \\
\left(\frac{2p+1}{5},5\right) & p \equiv 2 \pmod{5} \\
\left(\frac{3p+1}{10},10\right) & p \equiv 3 \pmod{5} \\
\left(\frac{p+1}{5},5\right) & p \equiv 4 \pmod{5} \\
\end{cases} \]](http://latex.artofproblemsolving.com/f/3/1/f31788496b757f35a9bad72566972493ef380dbf.png)
,
manually and assume 
.![\[\left\{-10, -\frac{p-5}{2}\right\} \in \left\{-2, -3, \dots, -\frac{p-1}{2}\right\},\]](http://latex.artofproblemsolving.com/d/c/f/dcf0ab3cdd34ff1d8ac4fd904e0b970970c70812.png)
but their reciprocals differ by![\[\frac{1}{10} - \frac{2}{p-5} = \frac{p-1}{2}\]](http://latex.artofproblemsolving.com/6/f/6/6f61f9ea21739c13b747e833f5f3ecab9889c7c2.png)
and thus cannot both lie in 
.
manually, and assume 
for some odd 
.
,
.
must be prime, both 
and 
are consecutive Fermat primes. Since 
and 
must be powers of 2, this forces 
,
.
,
.
must be prime, 
.
gives 
.
,![\[5 \mid 3 \cdot 2^{2^c-1} + 1 = p,\]](http://latex.artofproblemsolving.com/e/b/b/ebb17463b469c0e55e019a6233144705760eef73.png)
contradiction.
,![\[2^{k+1} \cdot \frac{p-n}{2} \equiv 1 \pmod{p}\]](http://latex.artofproblemsolving.com/7/f/a/7fa8f0447c3bf1c5d56a7c306bed42096d161011.png)
shows that this case yields no solutions.
be the smallest prime not dividing 
.
unless 
.
.
gives 
gives 
.
gives 
.
gives 
.![\[13^2 < \frac{2 \cdot 3 \cdot 5 \cdot 7 \cdot 11}{2}\]](http://latex.artofproblemsolving.com/5/b/9/5b98f220ce91778cb6e4e0299ef92216f6cb834b.png)
and 
for 
,
are the primes in increasing order. 
by minimality of 
,
.
,![\[\left(q \operatorname{rad}(k+1)\right) \cdot \frac{kp+1}{q \operatorname{rad}(k+1)} \equiv 1 \pmod{p}.\]](http://latex.artofproblemsolving.com/b/8/1/b81d2bfc75bfd2ab6b740cd18a5a7190a66e85eb.png)
If 
,![\[q \operatorname{rad}(k+1) < q^2 < \frac{p}{2}\]](http://latex.artofproblemsolving.com/8/6/d/86ded885ddc652c06e767ca1d78c18bd6ce54544.png)
and the second factor as![\[\frac{kp+1}{q \operatorname{rad}(k+1)} < \frac{(k+1)p}{(k+1)2} = \frac{p}{2}.\]](http://latex.artofproblemsolving.com/4/9/0/49077cf35734a27e1058e3e5db137b02863aa69e.png)
Therefore, no 
work. Indeed,
as desired.
.
if 
and 
otherwise. Thus, a prime 
for which 
.
are not mundane.
,
.
,
,
.
.
,
,
.
,
,
.
,
,
.
for which 
for some positive integer 
,
,
are divisible by 
.
,
are divisible by 
,
,
,
is not mundane.
,
,
.
are mundane.
which was not mundane (so, from now on, when we refer to 
,
.
.
,
.
.
modulo 
.
,
modulo 
as 
,
.
,
as well 
,
it follows that 
be the unique positive integer for which 
.
,
,
,
.
,
is a positive integer, 
,
contradicting the fact that 
,