AoPS Academy
, 
, 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
and 
are given. Let 
be the midpoint of the segment joining their centers, 
, 
be arbitrary points on , respectively such that 
. Find the locus of the midpoints of segments 
.
of points in the plane is balanced if, for any two different points 
and 
in , there is a point 
in such that 
. We say that is centre-free if for any three different points , and in , there is no points 
in such that 
.
, there exists a balanced set consisting of 
points. for which there exists a balanced centre-free set consisting of points.
and 
Prove that
such that p is a prime, 
and 
is a divisor of 
.
,
is 
or 
.
.
)
.
be the smallest prime divisor of 
.
.
.+p$](http://latex.artofproblemsolving.com/5/f/a/5fa0acf3aa516a7ef3e24f34f5a94863ee6d9c65.png)
,
,
,
,
.
.
,
,
(it it does satisfy the relation, since 
)
for any prime 
and 
.
.
and 
,
.![\[p^{p-1}|(p-1)^{p}+1=p^{2}\left(p^{p-2}-{p \choose 1}p^{p-3}+
\cdots + -{p \choose p-3}p-{p \choose p-2}+1\right)\]](http://latex.artofproblemsolving.com/5/d/c/5dcb92f1390234d96e36f164d0fd147d3ad72d49.png)
.
.
is odd, so is 
)
we get 
and 
.
(from the choice of q) leads to the existence of integers 
such that 
,
.
,
must be odd. This shows that 
and 
+p$](http://latex.artofproblemsolving.com/7/7/c/77c6492e0fa69aa7e7df2a126b7acbff9f9218ff.png)
,
such that p is a prime, 
and 
is a divisor of 
.
:
.
and 
.
.
has prime divisors later) 
always holds. So 
are also solutions.
.
be the smallest primedivisor of 
.
.
.
then 
.
.
when 
and 
.
.
.
gives the solution 
.
.
.
.
.
.
when 
.
.
:
.
.
and 
.
we see that 
is increasing when 
.
we see that 
.
we get the solutions 
we find that 
and 
,
.
,
.
but not by 
,
with 
and 
,
.
.
.
we also have 
.
we get 
.
it follows that 
,
.
which is clearly true, as 
,
is strictly decreasing for 
.
are the only solutions. 
,
,
.
.
and checking 
with 
.
.
for some integer 
.
.
for an odd integer 
.
.
and 
.
and 
have been dealt with using, for example, LTE (giving solutions of 
and 
)
.
)
.
and 
hence, as 
is twice a factor of 
is a factor of 
so 
.
is even). Thus 
.
so 
which is a case that has already been dealt with.
because 
and 
from fermat's little theorem,so 
,
Hence, the pair 
is valid for all primes 
If 
,
But note that 
,
It follows that 
,
,
such that 
By testing these six possibilities, we find the solutions 
If 
,
such that 
Thus we obtain the solutions 
Furthermore, note that if 
is odd, and so 
,
be the least prime divisor of 
Then we have the relation ![\[(p - 1)^x \equiv -1 \pmod{q^{p - 1}} \implies (p - 1)^x \equiv -1 \pmod{q} \implies (p - 1)^{2x} \equiv 1 \pmod{q}.\]](http://latex.artofproblemsolving.com/a/3/e/a3ee50dbee88492cabd08c1c8cc2d5f26ecbf45b.png)
Hence, if we let 
,
Furthermore, since 
and ![\[(p - 1)^{q - 1} \equiv 1 \pmod{q} \implies d \mid q - 1 \implies d \le q - 1.\]](http://latex.artofproblemsolving.com/f/f/b/ffb4e81cb71a68ef88ca3f8b02670e718760ef50.png)
Now, note that if 
,
that is also a divisor of 
)
It follows that ![\[(p - 1)^2 \equiv 1 \pmod{q} \implies p(p - 2) \equiv 0 \pmod{q} \implies p - 2 \equiv 0 \pmod{q},\]](http://latex.artofproblemsolving.com/b/5/c/b5c2670ace8ab326223bce7c9c787efcc27d9f7f.png)
because we assumed that 
It follows that 
However, this contradicts 
,
,
.
,
.
.
.
where 
and 
.
.
is odd 
)
because 
and the smallest prime that divides 
and 
.
or 
.
,
so we must have 
.
.
so 
,
.
.
.
)
.
and 
or 
.
)
.
,
.
.
.
,
.
.
)
.
)
.
.
.
.
the latter implies -1
Consider the smallest prime 
.
Then 
.
.
.
implying 
.
so 
works.
.
Then as we have alread considered 
,
.
and hence 
.
.
Which holds if and only if 
,
.
we have 
.
.
.
.
.
,
and as 
then 
so 
so 
,
impossible. If 
so 
so 
which by 
.
.
for the remainder of this solution.![\[\operatorname{ord}_q(p-1) \mid \gcd(2x,q-1) = 2.\]](http://latex.artofproblemsolving.com/1/0/8/108062964a700fa669f3c37f2182b2832e033196.png)
![\[(p-1)^2 \equiv 1 \pmod{q} \iff p(p-2) \equiv 0 \pmod{q}.\]](http://latex.artofproblemsolving.com/2/1/7/2170622b92e8835580d041c94c56b7b8037606ce.png)
.![\[(p-1)^x+1 \equiv 2 \pmod{q},\]](http://latex.artofproblemsolving.com/3/f/6/3f6552a8ea8199e5af45759d7eb88c6dbfa8f484.png)
.
,![\[\nu_p((p-1)^x+1) = \nu_p(p)+\nu_p(x) \ge \nu_p(x^{p-1}) = (p-1) \nu_p(x)\]](http://latex.artofproblemsolving.com/a/8/6/a8691725ff9b8220c2e4bb38bfe4be5c8121781a.png)
![\[\implies 1+\nu_p(x) \ge (p-1) \nu_p(x) \iff (p-2) \nu_p(x) \le 1.\]](http://latex.artofproblemsolving.com/1/a/3/1a3749942850b6c469d2934eafc9875706847b71.png)
,
.
.
.![\[(p-1) \cdot v_p(x) \leq v_p((p-1)^x+1) = v_p(x) + 1 \implies (p-2) \cdot v_p(x) \leq 1.\]](http://latex.artofproblemsolving.com/0/3/0/030dfcaefac6273de25d38ae1fe6152ceb145c66.png)
If 
.
.![\[\operatorname{ord}_q(p-1) \mid 2n, p-1 \text{ but not } n \implies \operatorname{ord}_q(p-1) = 2 \implies p=q,\]](http://latex.artofproblemsolving.com/0/0/f/00f9acaa793e4b6041504507e12d8f9f55e50f48.png)
contradiction. 
,
,
divides 
.
are relatively prime and thus 
,
,
and thus 
,
.
.
,
then they have equal values, so this case has to be resolved separately.
be the minimum other prime that is a factor of 
divides 
.
,
only contains prime factors greater than 
but 
has order 
.
.
which means 
which is either 
in which case we have 
which is divisible only by 
so 
then 
.
but if 
then that means 
.
.
,
clearly there are no solutions as the division no longer holds.
,
and 
Looking at the order of 
,
.
is impossible. To see why, notice that 
,
.
is not divisible by 
.![\[ \nu_p((p-1)^x + 1) = 1 + \nu_p(x) \geq \nu_p(x^{p-1}) = (p-1)\nu_p(x). \]](http://latex.artofproblemsolving.com/5/3/f/53f97bb18238ed1b38f680bae27299620fa5411d.png)
This implies![\[ \frac{1}{p-2} \geq \nu_p(x) \geq 1, \]](http://latex.artofproblemsolving.com/2/e/1/2e12b579e0888e705f156adb2f5ad7b9b3ee7308.png)
so 
and 
and 
as it always works, ignore 
as it can be done manually), let it be 
,
.
,
we can check that 
by size, so we can eliminate this case by just showing 
,
which forces no sol, since 
,
,
,
since the left is increasing in 
on the right is 
,
.
.
then 
and 
.
From this it's not very hard to show that 
.
then either 
or 
.
then by mod 2 we get a contradiction. If 
then by FLT 
so 
so 
is odd, so 
.
.
.
,
,
.
,
and 
.
.
.
,
so 
.
,
.
then 
.
,
then 
,
,
However, because 
,
,
.
,
and 
and 
,
,
.
and 
,
.