AoPS Academy
people who do not know each other. Prove that it is possible to introduce them in such a way that no three of them have the same number of acquaintances.
with circumcircle 
, excircle 
that is tangent to segment 
at 
. 
-mixtillinear incircle 
of 
is tangent to 
at 
. Suppose 
is the midpoint of the arc 
, and 
. Show that if the circles 
have the same radius, then tangents to at 
intersect on 
.
such that 
is a perfect square for all integers 
and 
.
such that for all 
the following holds:![\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]](http://latex.artofproblemsolving.com/a/c/a/aca098172ea7ce5663e9dd0dd10c2f1495630230.png)
be a prime number and 
an integer polynomial of degree 
such that 
and 
is congruent to 
or 
modulo for every integer 
. Prove that 
.
because 
is not identically 
(since 
) and has at least two distinct real roots, which means that 
.
is a suitable polynomial.
or 
. Maybe there's some stronger result and our new member did not quote the problem correctly.
, and regard the polynomial as being over 
. Now, assuming it has degree 
, because this is the interesting case, attempt to construct it using Lagrange's interpolation formula. The resulting polynomial will have the leading coefficient equal to 
, where 
is the number of residues among 
for which it takes the value . Since 
cannot be zero modulo , so the polynomial has a non-vanishing coefficient corresponding to its 
'th degree monomial.
![\[1^k+2^k+\ldots +(p-1)^k \equiv 0 \quad (\mod {p})\]](http://latex.artofproblemsolving.com/4/2/6/4267e739eeb8f90aba4394d6bead507e44b66ffc.png)
. Assuming that 
this yields![\[f(0)+f(1)+\ldots +f(p-1) \equiv 0 \quad (\mod p).\]](http://latex.artofproblemsolving.com/b/2/8/b28aadf7c55e8f20f74b8d0bcca614bc11fda275.png)
and 
, this is a contradiction.
and then from above information construct 
using 
...
to be 
.. A Contradiction to 
= ..
is a polynomial with integer coefficients and 
, then 
, each with 
is congruent to or , so the LHS is congruent to at most modulo , making it impossible to get a multiple of .
is clearly not a constant polynomial, so 
. Hence, 
is trivial.
. Assume for contradiction that 
.![\[f(x)=\sum_{j=0}^{p-1}f(j)\prod_{i\ne j}\frac{x-i}{j-i}\]](http://latex.artofproblemsolving.com/a/d/e/ade747a6a8a940524c2dde0a675e54ea16a36c17.png)
Note that is the unique polynomial obtained by Lagrange Interpolation, as ., the coefficient of in is zero (leading coefficient). So, we have![\[0=\sum_{j=0}^{p-1}f(j)\prod_{i\ne j}\frac1{j-i}=\sum_{j=0}^{p-1}\frac{f(j)(-1)^{p-1-j}}{j!(p-j-1)!}\]](http://latex.artofproblemsolving.com/7/e/7/7e7f09d2cc63fce4a8c2553103b208ae032594ca.png)
Since is even, we have![\[\sum_{j=0}^{p-1}(-1)^j\binom{p-1}{j}f(j)=0\]](http://latex.artofproblemsolving.com/e/b/d/ebd640cd06e6676f5ca28f6e2380cc3688d72523.png)
However, we know that 
![\[f(0)+f(1)+\cdots+f(p-1)\equiv 0 \pmod{p}\]](http://latex.artofproblemsolving.com/4/e/7/4e7ecaf16bf03d8d850bce351860bc729abd6562.png)
Since 
, it is forced that 
where 
.
and 
, we have that 
, which is impossible.
is true. 
![\[\sum_{i=1}^{p-1} f(i) \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/f/9/9/f9935a535f0ba50e73d1b646d31ecc94b9d10de9.png)
if 
which follows from the fact that 
for 
. must be constant modulo which is a contradiction by 
, so 
.
be prime. Find the smallest possible degree of a polynomial with integer coefficients for which 
, 
, …, 
leave exactly three distinct remainders upon division by .
is constructed by 
and FLT shows this works.
and the other value is 1. This is from deleting the constant factor and multiplying by the unit of whatever the other value is.
where 
Note that 
which must follow due to 
Hence, adding all of the values together yield 
However, the sum should be bounded(exclusively) by 
. Thus, a contradiction.
such that 
and define 
,
Q.E.D.
then 
,
