AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
is a triangle with circumcenter 
and 
is a circle tangent to 
and internally to 
The tangent
to intersects again at 
If 
are the contact points of with 
respectively, prove that 
+1 w, let 
be the excircle opposite to . Let 
and 
be the points where is tangent to 
, and 
, respectively. The circle 
intersects line at 
and 
. Let 
be the midpoint of 
. Prove that the circle 
is tangent to .
be an acute triangle. For a mobile point ![$M\in [AB]$](http://latex.artofproblemsolving.com/5/5/6/556a215c7a0922d0160f9e6c380d92ffecc7f57e.png)
denote 
. on the side ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
for which the distance 
is maximum.
be positive integers satisfying 
, 
. Suppose that 
are integers chosen from 
such that the set of remainders of the subset sums over all subsets of 
when divided by 
is exactly 
. Show that ![\[ a_1 + a_2 + \dots + a_k = l. \]](http://latex.artofproblemsolving.com/a/1/c/a1c908e4f0cffed380db8c0bcd0d4707ed59cdd1.png)
such that:
that satisfies 
for all reals 
and 
.
, such that 
and 
.
such that 
is prime, 
with 
and ?
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
such that:
be a prime number. Prove that there exists an integer 
with 
such that neither 
nor 
is divisible by 
.
has 
roots in 
. Assume what we're trying to prove doesn't hold. In each pair 
there's at least a root of , so there are at least 
roots of among the first 
residues 
. If is a root of , then so is 
(in , of course), so there are at least other roots of among the last residues . lie in the intervals 
and 
. However, we can clearly find roots which are not in these intervals, for example the product of two roots chosen from the pairs 
. Their product is 
and 
. For 
this product is outside the two intervals mentioned above, and for 
we can check it manually (take 
).
and 
. is equivalent to 
which is, after reducing 
,
is divisible by so, 
. So:
which means that 
and 
satisfies required condition. Indeed:
above congruence is not 
.:
. Let 
be the set of guys in 
such that 
, and 
be the others. Go by contradiction.
.
. We get the sum of a bunch of terms that are divisible by 
, then the last two are 
, which is not 1.
. into . Say 
. Then, . Now, consider 
, which is an injection since 
is prime (this is really just 
to 
). Then, 
, so 
. Thus, .
. are in . 
is even, so we can break up into the pairs 
. By assumption, each of these pairs has a number that, when raised to the power , is congruent to 1 modulo , so each pair has an element of 
. The conclusion follows.
. Now, note further that with our pairs from step 3, we in fact need exactly one member of each pair to be in . 1 is in , so 2 is not. This makes 3 in , so 4 is not. We see that we need the odd integers to be in , but the even integers to be in . We'll now proceed to get the contradiction.
. Observe that one of 
is not prime (one has to be divisible by 3, as is not). Both integers are odd, and thus have odd factors that are furthermore at most . If 
is composite, write it as the product of odd factors 
, and note that 
. The exact same thing is true of 
if it is composite. Now, we multiply by 
or 
, to get that 
or 
is also 1 modulo . Observe that is even, so these are the same as 
, which have even bases, and when , this means that they are not 1 modulo , contradiction.
. is not prime here, and so we can do the same thing as before with , to get 
not 1 modulo .
. 
. Oops.
. 
. Oops.
for 
.
.
, either or is 
mod . By our lemma, since 
, 
, so 
; repeating this argument, we find 
for 
; since is even, their negatives also satisfy the equation. Since 
is of degree , it can have at most roots mod , so these are all the roots. Thus, 
. Now the coefficient of the term 
is 
. But 
, the coefficient must be mod . Contradiction.
we have 
.
we r done because then we have
and so 
which is impossible because 
which means that we must have ij=1 which contradicts the fact that they are distinct odd natural numbers.
then we have that 
so we must check the case in which 
or better say 
for some n . we can check the case p=5 easily. so let's assume that n > 1. we have 5 | 2p+1 and so there exists an odd number distinct from 5 such that 
and it's easy to see that 
?
so there is no contradiction with Fermat theorem .
quadratic residue mod p if there exist two successif quadratic residue then take the first as ''a'' then we have ![$ a^{\frac{p-1}{2}} = 1 [p] $](http://latex.artofproblemsolving.com/f/5/0/f507e6958ff693ecd16b324296fd4818d00b57d6.png)
then by LTE we have 
and the same think for a+1 since it is also a quadratic residue . But still the other case can't be proven the same way .
modulo be "good" if 
, otherwise let it be bad. Then there are no consecutive bad numbers in 
, so at least half of them are good. If is good notice
is bad. Therefore there are no consecutive good numbers in either, otherwise if we take 
consecutive good numbers, then 
are consecutive bad numbers. Since every number is good or bad and 1 is good, we reach the conclusion that:
are all good, and so we have proven different numbers are good. Also notice that
modulo , there is only 1 good number. This implies there are good numbers in total, which MUST be the ones I just described (
and 
).
, we get that 
is NOT one of these numbers. Therefore, it is bad. But notice that 
is good, and so
, is good. Contradiction!
and 
and 
. The case 
is clearly impossible as 
, which can easily be verified through binomial expansion. Then, for some element 
, the elements 
and are all distinct, so 
. If 
, we are done as there must be two consecutive terms, so assume 
. If there are no two consecutive terms, since 
and , it must be that 
. Furthermore, 
since and 
are all distinct, so 
for , a contradiction.
in 
"good" if ![$a^{p-1}=1 [p^2]$](http://latex.artofproblemsolving.com/5/c/3/5c34ee02fc09c364645d0c70cbef748dc683282c.png)
. If the problem statement is false, then for every 
, at least one of and 
is good. Then at least 
numbers in 
are good.![$(p-a)^{p-1}=(a^{p-1}+pa^{p-2}) [p^2]$](http://latex.artofproblemsolving.com/a/e/e/aee16b23c96f8b1c78a813770e2d6832ced7dba3.png)
. If in and then clearly ![$(p-a)^{p-1} \neq 1 [p^2]$](http://latex.artofproblemsolving.com/f/2/1/f21b1c072e7859b81d651b46e8a73c8c2cd3afd1.png)
. Thus at most one of 
are good for every 
. Thus there are at most good numbers in , and hence there are exactly good numbers in . For the problem statement to remain false, it follows that are good and 
are not good. is good and 
is not good then 
is not good. In particular, if 
we get that since 
is good and 
is not good, 
is not good, a contradiction. If we can check that the problem statement is true manually.
. Hence for every 
there exist exactly 
values 
such that 
. Define 
, and note that 
. Clearly there are 
such . But for 
we have 
, and then it must be possible (by PHP) to find 
, as desired 
, show that for every sufficiently large prime there must exist consecutive integers 
such that none of 
is divisible by 
is an integer not divisible by , then 
iff 
(by binomial expansion) 
(because is odd) 
, then 
by FLT 
(because doesn't divide ) 
and 
for ![$a\in [\![1,p-2]\!]$](http://latex.artofproblemsolving.com/6/4/b/64b4f15954fe242fd338fb8d19b5277889bf4127.png)
: by our lemma, one can find that neither 
nor 
is divisible by , which would finish the proof 
divides neither 
nor 
so divides not 
so 
... and so each integer odd in ![$[\![1,p-1]\!]$](http://latex.artofproblemsolving.com/c/9/0/c90655e932324edfd1303e37cf485b27e4b4a646.png)
is such as and contrary for even . By our lemma, for an even a, because is even is odd, we get that 
, we have for 
, 
and 
, so that 
and then 
so 
, a contradiction 
has 
roots in 
"
But since in 
has exactly 
roots, 
has at most roots and 
has at most 
roots, has exactly roots.
.
has at most roots. Moreover, if is a root, then is also a root. Therefore there at most 
integers between 
and that are roots. Thus the only way there are no two consecutive roots is if all the odds or all the evens are roots. Since is obviously a root, it must be the odds. Thus![\[ x^{p-1}-1 = (x-1)(x+1)(x-3)(x+3) \cdots(x-p+2)(x+p-2) = (x^2 - 1)(x^2 - 9) \cdots (x - (p-2)^2 .\]](http://latex.artofproblemsolving.com/1/1/f/11f89ebbbca083849c7f6c68b2e7738978f9ebcd.png)
However, the coefficient of on the right hand side is![\[ - (1^2 + 3^2 + \cdots + (p-2)^2) = 2^2(1^2 + 2^2 + \cdots (p-1/2)^2) - (1^2 + 2^2 + \cdots (p-1)^2) = \frac{p (p-1)(p+1) - p(p-1)(2p-1)}{6} = \frac{p(p-1)(2-p)}{6} \]](http://latex.artofproblemsolving.com/2/4/b/24b9e4023ad2f08c5862d7e141d6678a569949e3.png)
which is not divisible by , contradiction.
consecutive numbers, at least one is divisible by . Also, note that 
always due to Fermat.![$p^3 | (i^{p-1}-1)((i+1)^{p-1} -1) \ \forall \ i \in [{1,2,...p-2}]$](http://latex.artofproblemsolving.com/7/7/5/77577e61ff92b83a7baae58c72fee5afd0470d4a.png)
.
can not divide 
. If this is true, then clearly the problem will be finished. Now first I tried this for 
, it works! Then I spent a (very) long time trying to play with that expression modulo but I could not do much at all.
hours spread out over two days. I tried to check this for 
but my calculator won't work upto that point (it just approximates the product then, which is not good for divisibility). So after working very long on this method, I gave up on this and worked through the official solution,
! But as I am not very experienced, I am not able to work through this, but these expressions seem calculatable mod but I just am not able to do it.
!
has precisely roots in 
and thus, 
doesn't have roots in 
Therefore, by Hensel's Lemma, for each 
there exists a unique 
such that 
is a root of 
in 
Consequently, has precisely roots not divisible by in It is obvious that 
is never a root of 
and thus, out claim is proven. 
has (at least) a root of in Hence, has at least 
roots in 
However, if is a root then so is 
roots in each of the intervals ![$[1,p-1]$](http://latex.artofproblemsolving.com/7/b/6/7b62614f5e864355c85377b7d30246f01d21efb1.png)
and ![$[p^2-p+1, p^2-1],$](http://latex.artofproblemsolving.com/b/0/6/b060b32a2d22d7293d3c407c82470d0c9d9be754.png)
and since is a root, then and must also be roots.
will also be a root, and 
and 
directly (both work with 
), we may assume 
. Observe that is impossible so let us consider only 
.
is a primitive root modulo and 
, 
, then 
if and only if 
, which is if and only if 
, i.e. divides . So the number of solutions to is 
which implies that the number of solutions in ![$\left[1,\frac{p^2-1}{2}\right]$](http://latex.artofproblemsolving.com/8/0/f/80f21d288ced4aeeee4b7af33b765c3e672f6f6a.png)
is (as if is a solution, then so is 
since is even). In particular, as is a solution and 
, the number of solutions in ![$[2,p-1]$](http://latex.artofproblemsolving.com/d/4/c/d4c08e8c3244e6b7d573f126d8c387866b225a76.png)
is at most 
.
, , among two consecutive elements at least one has to be a solution to . So the number of solutions in this set is at least , which equals the abovementioned upper bound. Equality is possible if and only if the solutions in ![$\left[2,\frac{p^2-1}{2}\right]$](http://latex.artofproblemsolving.com/5/a/2/5a22ec6c27f0db7f54aff06af3f35d301e0ec447.png)
are precisely , 
, 
, . Moreover, it is clear if 
and 
are solutions to , then so is 
- hence 
and 
is also a solution. However, 
and 
(equivalent to 
which is true for ) and this is a contradiction!
to be the set of such that 
. We would like to show that there are two elements of that are consecutive. Note that if 
(i.e. if 
), then 
It follows that in order for 
to be congruent to 
, we need 
, meaning that has to divide 
. But is coprime to , so 
if .
, so 
. Assume for the sake of contradiction that the problem statement is false. Then we must have 
, 
, and 
. If , then 
. Then, expanding and simplifying using the binomial theorem, we get 
., then 
. Expanding and simplifying using the binomial theorem, we get 
, which we can factor as 
Since , we must have 
, and so 
. So the only way that the above congruence could be true is if 
.
, so 
. Since the inverse of is 
, we now have 
. We see that in order for 
, we must have 
. Since we are given , we have now reached a contradiction, so it is not possible for both and , and we are done.
, and let be a generator of 
. The set of roots of are precisely the elements of the multiplicative subgroup 
. is a root iff 
is a root, the roots of the polynomial must also be 
for the problem to not be true., 
doesn't form a group under multiplication mod . This yields the desired contradiction. 
and 
. I claim that at least one of these works. doesn't work, i.e. one of 
and 
is divisible by . By the Binomial Theorem,
so we would require 
. Again by the Binomial Theorem, we have
Now we look at . We have
but since 
, this is impossible. Likewise, 
is equivalent to 
, which is also impossible, hence works if doesn't, so we're done. 
be the set of positive integers such that 
and 
We will first prove that at least one of and is in .. We have that 
which is impossible because 
, contradiction. and , inclusive, at most one of them is in . Specifically, 
must have exactly one element in . Assume for now that 
is the one not in (the other case is symmetric). Then, 
must be in , so must not be in . Then 
must be in , etc. This makes all odd positive integers within bounds in and all even positive integers within bounds not in , or vice versa. However, since is clearly in , we indeed have that all odd positive integers within bounds are in and all even positive integers within bounds are not in . and 
are both in . Plugging in,
Multiplying both sides by , we get that 
Similarly, we also have 
These combine to 
which means that 
, contradiction. QED.
-powerful if 
. If is -powerful and is -powerful then is -powerful. There are also at most -powerful numbers.
has a -powerful number for 
, and since 
is -powerful, the additive inverse of each of those -powerful is also -powerful. That gives numbers.-powerful:![\[\frac{(p-2)(p-1)}{2}\]](http://latex.artofproblemsolving.com/8/1/6/81664a84141ef2b3ad05640277b6dd3261456a47.png)
![\[\frac{(p-1)(p-1)}{2}\]](http://latex.artofproblemsolving.com/2/8/6/28650a8dfefd13a7d41f854bc792014e010546b4.png)
![\[\frac{(p-2)(p+1)}{2}\]](http://latex.artofproblemsolving.com/e/7/2/e720780bc7e2a05e004d3a465d4ac1fd9db1b758.png)
![\[\frac{(p-1)(p+1)}{2}\]](http://latex.artofproblemsolving.com/4/1/7/417977ef40c1eadb367db447e28e5a3291d97ec8.png)
but none of them are between and or 
and 
inclusive. Contradiction.
, , fortunate if 
and unfortunate otherwise. We wish to show that there exist two consecutive unfortunate residues mod . and is unfortunate. Suppose note. Then, 
and 
However, by binomial theorem the latter is 
However, since we know that 
is 1 mod , this means that 
is divisible by , which is clearly false. is unfortunate if and only if is even. and are both fortunate.
using binomial theorem would give us 
which rearranges to 
Doing the same with , we have 
If we square the first congruence, we get 
However, since 
is relatively prime to , 
and 
are both the inverse of mod . Hence, they are congruent mod , which means that 
clearly a contradiction as 
with 
, the terms in the binomial expansion of are all divisible by except 
and , so ![\[ (p-a)^{p-1} \equiv a^{p-1} - p(p-1) a^{p-2} \equiv a^{p-1} + p a^{p-2} \pmod{p^2} \ \ \ \ \ \ \ (1)\]](http://latex.artofproblemsolving.com/6/1/e/61e74f87f3bc74ba19e08cb284f1cc194fe87547.png)
, then 
. If 
(for 
) satisfied 
, then notice that 
would satisfy 
and 
aren't divisible by . Thus, exactly one of and 
is divisible by . Since 
, we have that 
so for each , 
iff is odd. (where 
) into 
, we see that the LHS is 
, so 
for any even 
. Multiplying both sides by gives that 
. Note that and 
are both at most 
, so 
and 
, so 
, so 
, absurd.
. Assume FTSOC such does not exist. Let 
be a subgroup of 
, where is a primitive root of . Note that 
. Since is even, if 
then 
. Hence![\[
|G\cap\{1,\ldots,p-1\}|=|G\cap\{-1,\ldots,-(p-1)\}|.
\]](http://latex.artofproblemsolving.com/4/9/c/49c970347349fc4bcc22582afd79b392e5a7067b.png)
By the assumption, we must have 
or 
(or both). Then 
or 
. Since , we have![\[
2^{-1},3^{-1}\not\in\{1,\ldots,p-1,-1,\ldots,-(p-1)\}
\]](http://latex.artofproblemsolving.com/5/7/7/577e6a0fa580cdb51169a0c02d72fe6a098ca44e.png)
so![\[
|G\cap\{1,\ldots,p-1\}|=|G\cap\{-1,\ldots,-(p-1)\}|<\frac{p-1}{2}.
\]](http://latex.artofproblemsolving.com/d/9/3/d933b611d3ed24d630d3dedd8b73bd0815bce71e.png)
Since there are pairs , it follows that one of these pairs satisfies the given condition, a contradiction. 
does not exist. Call an bad if 
. So out of ![\[ 1, 2, \dots, p-1 \]](http://latex.artofproblemsolving.com/e/b/0/eb07ec5e1b7617668f7119d4f2726842f392761b.png)
no two consecutive numbers may be bad. Hence there are at most bad numbers., and consider the polynomial ![\[ X^{p-1} - 1 = 0 \]](http://latex.artofproblemsolving.com/1/f/4/1f4d611cb839cc625eebe4c736fb4925aa98907a.png)
in are precisely![\[ \{ 1, 3, 5, \dots, p-2 \} \cup \{ p^2-p+2, p^2-p+4, \dots, p^2-1 \}. \]](http://latex.artofproblemsolving.com/8/6/e/86e2628c2fd1d3b09f4921426eef06efd4740489.png)
Proof. It is well known, due to Frobenius, that 
has roots in . bad numbers, then there are at least roots in 
. is a root so is 
. Thus there are also at least roots in 
. bad numbers. Since they cannot be consecutive, and is not bad, it follows that 
are the bad numbers. Thus 
are good. ![\[ 1 = 3^{p-1} \implies \left( \frac 13 \right)^{p-1} \equiv 1 \]](http://latex.artofproblemsolving.com/4/9/1/491bfae14549494340cfefe1b94a8422f5e3bbac.png)
But 
, since 
.![\[ \frac{2p^2 + 1}{3} \le p-1 \implies 2p^2 - 3p + 4 \le 0 \implies p \le 2 \]](http://latex.artofproblemsolving.com/b/a/b/babaebf8864337fc5f0fac45e7f498826cb755f0.png)
absurd, and ![\[ \frac{2p^2 + 1}{3} \ge p^2-p+1 \implies 0 \ge p^2-3p+2 \implies p \le 2 \]](http://latex.artofproblemsolving.com/4/8/2/4826e9cdd18e7b0b6ddb50d800b2af56115bf65b.png)
absurd again. Thus 
, so we found another root, contradiction. 
is good if and only 
and bad otherwise. The following is the key claim., at most one of and is good. to be good. Then,
since 
, which clearly implies that 
is bad. (consider ) and (consider again ). Now, we assume that there exists some prime 
such that the desired condition is not satisfied. Then, one of 2 or 3 must be good. If 2 is good, is not good, which implies that 
must be good. This in turn implies that is not good, and so on. Thus, if is good, all 
such that 
must be good, and the rest must be bad. Similarly, if is good, all 
such that 
must be good, and the rest bad. is good, note that one of and 
must be a non-power of two (). Thus, we can write this integer in the form 
for positive integers and 
. But then,![\[(2^kl)^{p-1} \equiv 1 \pmod{p^2}\]](http://latex.artofproblemsolving.com/1/8/9/1892af57683918f0487619173c25ef091251d620.png)
and![\[(2^k)^{p-1} \equiv 1 \pmod{p^2}\]](http://latex.artofproblemsolving.com/9/9/8/9981b58f8e62a7b955181144e17161c5385f1102.png)
But this implies,![\[l^{p-1} \equiv 1 \pmod{p^2}\]](http://latex.artofproblemsolving.com/8/1/c/81c99372680cfdca95e2dfb590366236b368533d.png)
which is a clear contradiction. is not good but is good. Note that then,![\[2^{p-1}-1 \equiv \frac{p}{(p-2)} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/c/9/ac9918e033f130fd0c51f561644f9a27e1073fe5.png)
and![\[6^{p-1}-1 \equiv \frac{p}{p-6} \pmod{p^2}\]](http://latex.artofproblemsolving.com/e/9/1/e91a38d4544cf78122fc261eb31c710b97b1ccce.png)
Thus,![\[1 \equiv 3^{p-1} \equiv \frac{\left(\frac{2p-6}{p-6}\right)}{\left(\frac{2p-2}{p-2}\right)}\]](http://latex.artofproblemsolving.com/7/6/b/76b9c674429dd2c757aee07f2bcd2f4e29c587ac.png)
which implies,![\[(p-3)(p-2) \equiv (p-6)(p-1) \pmod{p^2}\]](http://latex.artofproblemsolving.com/b/b/3/bb353c26ec571ebe3600b4f37b1a76dd74da2818.png)
Thus, 
which is a contradiction for all primes 
. Thus, this case is impossible too, indicating that our assumption that there exist no two consecutive good integers must have been false, which finishes the proof.
, we choose , therefore we can assume 
we have 
such that neither
nor 
is. Let 
then note that
or 
is not divisible by . for which there exists no such s.t. both and 
are not divisible by 
for all 
.Now if for some 
both 
and 
are divisible by then this implies both of 
and 
are not divisible by which is a contradiction, hence only one of and are divisible by . Hence the divisibility by alternates and as we have 
, this implies 
for all odd s.t. 
. As, we get 
for all even s.t. . Therefore we have
, hence 
and 
, hence 
and 
, hence 
. So we have
, hence
so 
which is a contradiction.
and 
Prove that 
and 
Prove that 
implies 
.
,
.
when we don't know a lot about m.
.
.
,
are in 
,
.
,
.
,
.
by our lemma, and 
,
,
are distinct 
If 
which is distinct than above 
Since, 
lies among those 
and if 
both are solutions then so is 
.
solutions must lie in the interval 
and others in 
.
we have one of 
or 
divisible by 
to denote 
wherever required. Now we proceed with the following claim.
Proof. Using ![\begin{align*}
(p-k)^{p-1}-1 &=\left[\overbrace{\binom{p-1}{0}p^{p-1}-\binom{p-1}{1}p^{p-2}k + \ldots}^{\text{Rest are}~0 \pmod{p^2}} - \binom{p-1}{p-2}pk^{p-2} +\binom{p-1}{p-1}k^{p-1}\right] -1 \\
&\equiv -(p-1)pk^{p-2}+k^{p-1} - 1 \pmod{p^2} \\
&\equiv -pk^{p-2}+k^{p-1} - 1 \pmod {p^2}\\
\end{align*}](http://latex.artofproblemsolving.com/b/1/7/b1789e6926c7ef253cdd7d5df6ea59457d2fde2b.png)
as desired. 
in previous claim, we get 
but 
.
whence 
due to assumption but this implies 
whence we get 
since otherwise 
a contradiction. Previous result immediately implies, via assumption, that 
which gives 
Continuing in the stated manner, we get the alternative sequence as 
Therefore, we get![\begin{align*}
&\iff(p-2)^{p-1}-1 \equiv (p-4)^{p-1}-1 \\
&\iff -p2^{p-2}+2^{p-1} \equiv -p4^{p-2}+4^{p-1} \\
&\iff p2^{p-2}[2^{p-2}-1] \equiv 2^{p-1}[2^{p-1}-1] \\
&\iff p[2^{p-2}-1] \equiv 2[2^{p-1}-1] \\
&\iff 2^{p-2}[p-4] \equiv p-2 \\
&\implies 2^{(p-2)(p-1)}[p-4]^{p-1} \equiv [p-2]^{p-1} \implies 2^{(p-2)(p-1)} \equiv 1 \pmod{p^2} \\
\end{align*}](http://latex.artofproblemsolving.com/d/1/6/d16ac8624239ae818ecfcdc3c9ed12f606cb294a.png)
but now notice that 
.
then 
but here it's a contradiction (here, not in problem now) since 
.
where 
.
which gives 
which is a contradiction. Therefore, our result holds, i.e. the problem statement. 
,
,
.
and we get,
So, 
Finally we slve a system of equations,(Consider them in modulo 
Thus we have 
,
We claim 
Indeed, 
Suppose FTSOC that at most one of 
is an element of ![$a\in[1,p-2].$](http://latex.artofproblemsolving.com/3/8/a/38ade32f29bb10e6002c665dba2964d6aec4f680.png)
Notice 
Hence by 
and 
Therefore, 
or 
or 
since 
This is absurd, as it implies 
and we have 
