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such that 
.
be positive integers and 
be prime. We know that 
and 
. Prove that 
is a perfect square.
be a set of 
residues 
. Prove that there exists a set 
of of residues such that 
contains at least half of all the residues .
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the -th derivative of at 
) is divisible by 
order, and if the constituents of its first, second, ..., rows are the first 
figurate numbers of the first, second, ..., orders respectively, show that it's value is 
be expressed as a polynomial in 
with real coefficients?
holds inequality 
for which there exists positive integers 
, 
, and 
satisfying![\[
2a^n+3b^n=4c^n.
\]](http://latex.artofproblemsolving.com/0/2/9/0295405b01494edc05ce9867b2ec48a4e069bf26.png)
such that following equality holds
by 
, and![\[ a_{n+1} = \left(n+1-\gcd(a_n,n) \right) \times a_n \]](http://latex.artofproblemsolving.com/1/3/1/1316c6496b6faec9bbaab8d87560bf995ad2141d.png)
for all integers 
.
if and only if is prime or 
.
denotes the greatest common divisor of 
and 
.![\[ \dfrac {\text {gcd}(m, n)}{n} \dbinom {n}{m} \]](http://latex.artofproblemsolving.com/c/0/a/c0a6877cd49e904728b24e7743fe8021efd6d312.png)
is an integer for all pairs of integers 
.
does not imply a = b = c.
, 
and 
be the centers of its A-excircle, B-excircle and C-excircle, respectively. Then, we have
,
,
. is the center of this A-excircle, 
, so the triangle 
is right-angled at T; also, 
(since the point , being the center of the A-excircle of triangle ABC, lies on the angle bisector of the angle CAB), and AT = s, where 
is the semiperimeter of triangle ABC. Thus, in the right-angled triangle , we have 
, so that 
. But by the half-angle formulas, 
. Thus,
.
and 
. Hence, we can equivalently transform the relation as follows:
.
and similarly 
and 
, this becomes
.. Then, we must therefore have . Now, since our situation is symmetric, we can WLOG assume that 
. Since 
, we have
.
;
becomes 
, i. e. 
. Division by a transforms this into 
, and thus
.
, what is a quadratic equation in a and has the solutions and 
, of which the second one can be excluded since it is negative (and sidelengths of a triangle cannot be negative), so we get the solution b = c and . The other solutions can be obtained similarly (we WLOG assumed that and thus didn't get the cyclic permutations). By tracing the above argumentation backwards, we see that these solutions are indeed solutions.
are equally, i.e. 
. Thus,
, what is absurd. Therefore, 
a.s.o.
be a function such that 
.
.
s.t.![\[ AI_a : BI_b : CI_c = BC: CA : AB \]](http://latex.artofproblemsolving.com/0/9/3/09318ace664b6b05a258fba9f8c6cbc815b895d2.png)
where 
where 
is the radious of the excircle!!
.
