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.
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
+1 w and are primes and is a perfect square, prove that 
is a perfect square.
be a positive integer. Aslı and Zehra are playing a game on an 
chessboard. Initially, 
stones are placed on the squares of the board. In each move, Aslı chooses a row or a column; Zehra chooses a row or a column. The number of stones in each square of the chosen row or column must change such that the difference between the number of stones in a square with the most stones and a square with the fewest stones in that same row or column is at most 1. For which values of can Aslı guarantee that after a finite number of moves, all squares on the board will have an equal number of stones, regardless of the initial distribution?
are real numbers, find the smallest possible value of the expression:
reals, and 
.
?
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
such that: Among every vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
such that a positive integer power of 
equals to a positive integer power of 
.
be an acute-angled triangle with 
A circle passes through 
and 
and crosses the segments 
and 
again at 
and 
respectively. The circumcircles of 
and meet each other at points 
and 
The segments 
and 
intersect at 
Let 
and 
be the reflections of 
in the lines 
and 
respectively. Prove that the points 
and 
are concyclic.
be positive real numbers such that 
. Prove that
When is the equality valid?
be a center of circle which passes through vertices of quadrilateral 
, which has perpendicular diagonals. Prove that sum of distances of point to sides of quadrilateral is equal to half of perimeter of .
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
cards on the table, and each card has a positive integer written on it. An odd number is written on exactly 
cards. Every minute, the following operation is performed: for all possible sets of 
cards on the table, the product of the numbers on these three cards is calculated, all the obtained results are summed, and this sum is written on a new card and placed on the table. A day later, it turns out that there is a card on the table, the number written on this card is divisible by 
Prove that one hour after the start of the process, there was a card on the table that the number written on that card is divisible by
on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
and 
written on the board, erasing them, and writing the integer 
. and written on the board, erasing them, and writing the integer 
.
is called a winning number if there exists a sequence of operations such that, at the end, the only integer left on the board is . Find all winning integers among 
and, for each of them, determine the minimum number of GCD operations Charlotte must use. denotes the greatest common divisor of and , while the number denotes the least common multiple of and .
and such that 
divides 
is the only solution. If 
one can easily check that there are no solutions so suppose 
. is odd; this gives 
. Take 
an odd prime dividing . Then divides . As 
and 
are coprime, we get that divides 
, where 
. Thus, divides 
so the order of mod has to divide 
. If the order is 
or ), then divides both 
and , so divides 
, a contradiction. If the order is , as divides , we get that divides 
so divides 
and hence 
. Otherwise the order is so divides 
. Thus, if 
for some odd distinct primes 
and some nonnegative integer 
(which by LTE is less or equal to ), by the previous we have 
so 
and thereby we get 
. A quick check gives 
the only possibility.
, we have 
divides . If is not a power of two, proceeding as in the previous case we get that any odd prime divisor of must be congruent with mod . As is even, we get that 
which gives again 
so , a contradiction. Thus 
for some positive integer . Clearly has to be even, 
so 
. As 
and 
we get that either 
or 
. We therefore infer that either (in which case again but we supposed ) or 
. However, and would give 
dividing 
, a contradiction.
be an odd prime dividing , so consequently divides . ( if is odd we have that 
, and since this number is clearly bigger than 
, we know that it must have an odd prime divisor, and if is , our number is odd, and so our hypothesis is obvious) is we verify by hand.
, so 
. Let the order of 
be 
. Thus, divides both and and, since it can't be odd, it has to be either or . If it is , we then have that divides 
and, since is odd and not , it has to be 
. If is , we have that 
. Taking these into consideration, we get that 
, where 
and 
. By LTE, we have that is if 
(we verify by hand), or 
. Also, 
.
, we get that 
, so 
. Thus, and, after some casework, . Thus, the only solution is 
.. If there exists a prime that divides and is not , we proceed as previously. if is a power of , then it's exponent has to be even.
, and so is or , which means that is either or 
. We get a contradiction for both cases. divides , then 
. 
is either or 
, and 
is . Thus, 
, and we verify the cases by hand.
?
yields that
? yields that
only which works. We first dispense of several edge cases:
has no solutions. has only the solution . is either , , or 
. which divides . Evidently 
. Apparently, 
, so either 
meaning 
, or 
has order , as needed. 
is the product of , , and primes. We now consider some new cases:
and 
, then , while 
. Thus, we must have ![\[ 3p^{q-1} + 1 = 2^2 \cdot 7^{0\text{ or }1} \cdot \left( \text{$1 \bmod p$ primes} \right). \]](http://latex.artofproblemsolving.com/a/5/8/a589e936e08032a0188714d22e029ed10b59d44d.png)
This gives that either 
or 
, but this gives 
.
, the same analysis holds except that the exponent of must be exactly equal to 
(despite 
)., the same argument works except that we need the additional observation that 
; hence ![\[ 3p + 1 = 2^{1\text{ or }2} \cdot 7^{0\text{ or }1} \cdot \left( \text{$1 \bmod p$ primes} \right). \]](http://latex.artofproblemsolving.com/6/5/0/6505d329221db72da3d3812c008aaa8b69e3e8a0.png)
Hence we have one extra case 
to check, which fails.
is the only solution.
and such that divides 
there are no solutions. We begin like v_enhance, so if 
, then 
or 
. Looking at 
, there are no solutions so 
is maximum value of 
and similarly 
. Now, annoying casework gives 
as the only solutions.
? yields that?
only, which clearly works. We can verify that for no solutions exist, and for we only have . Henceforth assume 
We now begin with the following important claim: divides , then 
or 
. be such a prime. Clearly 
. Then we have
For convenience, let 
. Then we have 
but 
, so either 
or 
. If the former is true, then we have 
, which gives 
. If the latter is true, then since 
, we have 
, which is the desired conclusion. is a divisor of , it follows that is the product of 
and primes 
. Further, we have
as is odd, so 
. Now suppose that 
. In this case, 
is even, so 
, so we require 
. Further, by exponent lifting we have 
, so it follows that
where is the product of primes that are and 
—this is because if 
we have 
and if we have 
anyways. Note that 
. Now taking 
, we have 
, which is either or . But we also have 
, so either 
or 
, giving . Since we supposed this case gives no additional solutions, then we need 
. Similarly to the previous case, if then , otherwise we have 
, so
where is the product of primes and 
. LIke the previous case, this means that at least one of 
is 
, but these respectively imply that 
. Since , the only possibility is therefore 
. But we can manually check that 
fails, so is the only solution. 
be any prime dividing 
. Clearly 
. Hence, we can write 
so that 
. It follows that 
, so either 
or 
. We have several cases.
. We can manually verify that does not work, so is odd. Moreover, clearly 
does not divide , so the only prime divisors of are and . If 
, then 
so 
by LTE, a contradiction. Hence, 
. But since is odd, 
. Combining these results, we obtain 
. However, the only solution to this is , which we have already checked.. We can manually verify that there are no solutions in this case. and are odd and consequently 
.. We can manually verify that the only solution in this case is 
.. Then does not divide , so 
, a contradiction.
and 
. By LTE, 
. It follows that either 
or 
is equivalent to mod , a contradiction as 
.
.
's in the proof are the actual fraction, not the value of the Legendre/Jacobi symbol. be a prime that divides . Then 
or .
obviously fails, so ![\[r\mid\left(\frac{11}{17}\right)^p+1.\]](http://latex.artofproblemsolving.com/a/4/6/a46ca76dec8cfb84b871914f7a8f974c281c22ba.png)
If we let 
, we have 
. Also, we have that 
unless 
(which is one of the cases anyway), so 
. If 
, then ![\[r\mid \left(\frac{11}{17}\right)^2-1\rightarrow r\mid 121-289\rightarrow r\mid 168\rightarrow r\in\{2,3,7\}.\]](http://latex.artofproblemsolving.com/c/a/d/cad520d8282fc9854d95aee6a4ba3bd9dd31a2d0.png)
But 
would imply 
, so . Otherwise, 
, so 
, completing our proof.
, then 
.
. Since is odd, 
is a square, so let 
and 
. Taking mod 
, we have![\[\nu_2\left(3b^2+1\right)=2.\]](http://latex.artofproblemsolving.com/2/6/4/264d4fcecf6bd3c2c10751d3132f9ba099190c18.png)
Also,![\[\nu_7\left(3b^2+1\right)\le \nu_7\left(11^p+17^p\right)=1+\nu_7(p)=\begin{cases}1&p\ne7\\2&p=7\end{cases}.\]](http://latex.artofproblemsolving.com/b/6/5/b653df52192d9bfcc1440aa0924efe8ee93127d5.png)
Thus 
has to be equivalent to at least one of 
, 
, and 
mod . So 
, 
, or 
. Since 
, that case doesn't work, so , which is what we wanted., then 
and 
.![\[\nu_2(3p+1)\le\nu_2\left(11^p+17^p\right)\le\nu_2(11+17)+\nu_2(p)\le 3\]](http://latex.artofproblemsolving.com/9/c/3/9c384c0c4235310725fd54af3156e97f011e6a56.png)
![\[\nu_7(3p+1)\le\nu_7\left(11^p+17^p\right)\le\nu_7(11+17)+\nu_7(p)\le 2.\]](http://latex.artofproblemsolving.com/5/2/5/5255d87d33369b61d9541d48f20db126c3fd4a34.png)
: Then 
, so ![\[3\cdot 2^{q-1}+1\in\{1,2,5,10,41,82,205,410\}.\]](http://latex.artofproblemsolving.com/4/d/2/4d2d3b0537cba9a5cb4c4889da61c97fdb2b407c.png)
Then ![\[2^{q-1}\in\{0,3,27,68\},\]](http://latex.artofproblemsolving.com/8/7/1/871f6ff607ab49d8519c6b2472d38d5b454a2b3f.png)
so there are no solutions in this case.: Then![\[11^3+17^3=(11+17)\left(17^2-11\cdot 17+11^2\right)=28\cdot 223.\]](http://latex.artofproblemsolving.com/e/6/f/e6fb5b46cb8824420522f02f91e45e82a5f05466.png)
Therefore,![\[3^q+1\in\{1,2,4,7,14,28,223,446,892,1561,3122,6244\}.\]](http://latex.artofproblemsolving.com/7/b/e/7be6b1b6aff33a505122a49c892154ec82c4d722.png)
We can see that![\[3^q\in\{0,1,3,6,13,27,222,445,891,1560,3121,6243\}\rightarrow 3^q\in\{3,27\}.\]](http://latex.artofproblemsolving.com/c/e/c/cec41cc98deb84fdeef651627585cdc5ea9d5994.png)
But 
is not a prime, so only works here.: Then ![\[3p+1\mid 11^p+17^p.\]](http://latex.artofproblemsolving.com/2/d/4/2d4cad11f4dd313d8af88c6ed543f95b9abcb418.png)
By lemma 1, all prime factors of are either , , or mod . In the third case, 
is too big, so 
, absurd. So![\[3p+1=2^m 7^n\]](http://latex.artofproblemsolving.com/9/e/b/9ebda04ba16fa8bf8edbd5b753e12bf95723a117.png)
for nonnegative integers 
where 
and 
(by lemma 3). Therefore,![\[3p+1\in\{1,2,4,8,7,14,28,56,49,98,196,392\}.\]](http://latex.artofproblemsolving.com/3/4/d/34d2df3fa6720fd89343f236d0e99ed707e49cc6.png)
This implies that![\[p\in\{0,1,2,9,16,65\},\]](http://latex.artofproblemsolving.com/9/1/3/91380ab6c07e12185219f38618bebc15bdc97b7a.png)
all of which are impossible.
, which obviously works. QED.
be a prime that divides . Then as 
, either 
or 
.
. Notice that if 
are both odd primes, then 
precisely. On the other hand, is not a quadratic residue mod , so this means that 
. But this means now that 
as 
so , and correspondingly too., then note that 
and 
, which upon a finite case check yields no solutions.
is the only solution.
and are odd primes. Note that and hence due to size reasons we must have an odd prime factor . Now consider a prime factor of 
, say 
. Then we have,
Then we have 
. Then either 
and hence 
, or we have 
. Then all prime factors of are , or modulo . As a result all prime factors of are equal to or or are congruent to modulo . Now note that 
as if it is easy to see that 
else if , then 
. Assume 
. However then,
However as is a NQR modulo we cannot have . Thus 
where is the product of primes congruent to modulo . However then obviously we require 
or 
and hence . Now we then have,
from which we find . Now if we find,
Note that from our claim the primes dividing are , or are modulo . Clearly 
and we require 
and 
. Hence 
and a finite case check leads to no solutions.
, which clearly works. be a prime that divides . Either , , or 
. obviously cannot be , and clearly works, so assume 
for the remainder of this proof. We write![\[r \mid \left(\frac{11}{17}\right)+1,\]](http://latex.artofproblemsolving.com/d/0/4/d0432226eaeacb7ee0c66ff6fcfa0f12da8ce968.png)
modulo divides ; denote this order as 
Furthermore, does not divide , so can be or . If 
,![\[\left(\frac{11}{17}\right)^2 \equiv 1 \pmod{r} \implies -168 \equiv 0 \pmod{r} \implies r \in \{3,7\}.\]](http://latex.artofproblemsolving.com/e/3/6/e3667a6f3988b76a5e88cbaa94d1d5e32fa5da13.png)
, so is the only extra case here.
, we have 
, which gives the last solution set. 
, then .. Let be the positive integer such that 
. Notice that![\[3a^2+1 \equiv 4 \pmod{8}.\]](http://latex.artofproblemsolving.com/6/e/a/6eafbfa7251fb7286b19ef1288308fbb7556e6d9.png)
![\[\nu_7(3a^2+1) \le \nu_7(11^p+17^p) = 1+ \nu_7(p) =1.\]](http://latex.artofproblemsolving.com/7/2/f/72fc8312a042baaa517cf30e6e71ff4cd17521b2.png)
![\[3a^2+1 \equiv 2^2, 2^2 \cdot 7 \pmod{p}\]](http://latex.artofproblemsolving.com/8/5/3/8538cc8aa630545956814247099eec00d683c0fb.png)
![\[\implies 1 \equiv 4 \pmod{p} \ \textbf{ or } \ 1 \equiv 28 \pmod{p}.\]](http://latex.artofproblemsolving.com/d/7/5/d75c42a62350006425e259441fbd5839938c74a0.png)
. If , we must analogously have![\[1 \equiv 196 \pmod{p},\]](http://latex.artofproblemsolving.com/f/b/f/fbf4927b817a1cc58eb209f34ad57368fde9acf6.png)
, we have![\[3 \cdot 2^{q-1}+1 \mid 410\]](http://latex.artofproblemsolving.com/b/3/d/b3d7febbc39856b9bd4a1b4666958a4b00ae1091.png)
![\[\implies 2^{q-1} \in \{0,3,27,68\},\]](http://latex.artofproblemsolving.com/3/3/0/330b0a17f387f3d3115c90eae0d7e934ace36b28.png)
, we have![\[3^q+1 \mid 6244\]](http://latex.artofproblemsolving.com/d/4/b/d4b754d5461b9facfe26b015bdd6859e670c9bc9.png)
![\[\implies 3^q \in \{0,1,3,6,13,27,222,445,891,1560,3121,6243\}.\]](http://latex.artofproblemsolving.com/d/2/b/d2b9b4c7ae2d455566c29d7345adc960327ec543.png)
., we have![\[3p+1 \mid 11^p+17^p.\]](http://latex.artofproblemsolving.com/3/8/f/38f18450ff2a1fbdc7ad4cf12090b4146a025b17.png)
are , , or 
for some positive integer . Clearly, the latter case is out of the question, so we have that![\[3p+1 = 2^m7^n.\]](http://latex.artofproblemsolving.com/6/4/1/641aea808d9a1b13fc6bde4d40dccd522f2c367c.png)
![\[\nu_2(3p+1) \le \nu_2(11^p+17^p) \le 3\]](http://latex.artofproblemsolving.com/d/6/a/d6a158f706c0ff939c1db971021171b0eeff039c.png)
![\[\nu_2(3p+1) \le \nu_2(11^p+17^p) \le 2.\]](http://latex.artofproblemsolving.com/7/e/5/7e5b4dd1cc7e6c86db8e1b90bc17e13282bdc288.png)
, which yield
, 
, and 
later, so assume none of these are true.
is a prime such that 
, then 
or 
.![\[ r \mid 3p^{q - 1} + 1 \implies 11^p + 17^p \equiv 0 \mod{r} \implies \left(\frac{-11}{17}\right)^p \equiv 1 \mod{r} \]](http://latex.artofproblemsolving.com/5/c/e/5cea171bd7d7675132e33748f90ac19eaceacdc1.png)
be the order of 
. We have 
. If 
, then![\[ r \mid \frac{-11}{17} - 1 \implies r \mid 28 \implies r = 2, 7 \]](http://latex.artofproblemsolving.com/3/4/7/347c77125c9a299d1f1e6e261d7ac58e7d265fdb.png)
, we have 
.
is even and 
is odd,![\[ p^{q - 1} \equiv 1 \mod{8} \implies 3p^{q - 1} + 1 \equiv 4 \mod{8} \]](http://latex.artofproblemsolving.com/b/d/a/bda225836e5508499c52c5afb82a041804e957ad.png)
![\[ 3p^{q - 1} + 1 = 2^2 \cdot 7^\alpha \cdot p_1^{\alpha_1} \cdots p_k^{\alpha_k} \]](http://latex.artofproblemsolving.com/9/b/0/9b03745615d5abba8b5e703a7e2de61a1528cb22.png)
.
, 
![\[ v_7(11^p + 17^p) = v_7(28) + v_7(p) = 1 \]](http://latex.artofproblemsolving.com/4/c/2/4c24d612e3e98f20a90304356d00ffffd5f5ff73.png)
, and![\[ 2^2 \cdot 7^\alpha \cdot p_1^{\alpha_1} \cdots p_k^{\alpha_k} \equiv 28, 4 \equiv 1 \mod{p} \]](http://latex.artofproblemsolving.com/b/8/9/b894403f09117996f4d6fea29baec5c01dddbaed.png)
.![\[ 3^q + 1 \mid 11^3 + 17^3 = 6244 = 7 \cdot 4 \cdot 223 \]](http://latex.artofproblemsolving.com/6/9/0/69057b0474ef00f05dae167dbac29b8af288985f.png)
., we have![\[ 3 \cdot 2^q + 1 \mid 11^2 + 17^2 = 410 \]](http://latex.artofproblemsolving.com/0/8/3/083bbedada9dd0637e07f0fd3b14d8e2286bea61.png)
, we have![\[ 3p + 1 \mid 11^p + 17^p \]](http://latex.artofproblemsolving.com/f/5/d/f5d97f5e0e3f098b9b388e67b9f77d34ca8800da.png)
, which means 
, or 
, and , which doesn't work.,![\[ v_7(11^p + 17^p) = 2 \]](http://latex.artofproblemsolving.com/1/2/2/122ad5ecf1d1412fc1f90c0f027b9d2c628fb432.png)
can be 
, where 
yield the same cases as earlier and 
yields![\[ 196 \equiv 1 \mod{p} \implies p \mid 195 \]](http://latex.artofproblemsolving.com/d/0/1/d01007cddd0d92a527b99b07da025f7ff6e13b26.png)
.
.
be a prime factor of . Then we have that either 
or 
be all the prime factors of we have that all 
.
then we have that the term with in the product on RHS is even while on LHS it is odd contradiction.
can be the prime factors of , then a case bash gives as the only solution.
such that it divides so 
or , so 
or 
.
then we get 
which is not possible.
. We check for which gives us 
which is not possible. For we have 
, after case-bash we get .
then we have 
, if 
then 
. So 
, this gives us to be a solution. we have 
, since cannot be written as 
, so it is of the form 
, also we can write it as and because 
we have 
, which does not work..
be a prime dividing , then we get by taking the inverse 
congruent to 1, giving either being or 2p divides r-1, for the second case, is congruent to modulo 2p, which is impossible for p odd prime. Using LTE on for , we get that, the most power is 1 and can never divide , this means that 
where 
, check the remaining cases.
which divides , so 
and thus 
or , giving or .
. Then 
(by mod ) and 
(by Lifting the Exponent) which is if and if . So (for actually it is ) and 
Hence ![\[ 3p^{q-1} + 1 = 2^{\leq 2}7^{\leq 1} \
\cdot \ (\mbox{primes } \equiv 1 \pmod{2p}). \]](http://latex.artofproblemsolving.com/5/4/c/54cf4bfab52d828594b1d6800bc40c8c3e1f251b.png)
Taking the latter mod yields 
. Since 
, checking 
gives no solution. For 
checking 
gives , so is a solution. If 
, then the right-hand side above must be divisible by , but then mod yields 
, contradiction for any positive integer . is prime!
, which works.
satisfies 
or . be a prime divisor of 
and 
.![\[11^p + 17^p \equiv 0\pmod r\]](http://latex.artofproblemsolving.com/9/9/d/99d4591fd85dbef282b049ec10ac85f57b807592.png)
![\[\left(\frac{11}{17}\right)^{p}\equiv -1\pmod r,\]](http://latex.artofproblemsolving.com/8/c/3/8c34782959dfca007dbe51d727a6a984dd1aa64b.png)
so ![\[\left(\frac{11}{17}\right)^{2p}\equiv 1\pmod r\]](http://latex.artofproblemsolving.com/6/5/3/6535582ca500db0ce89e14a2730d18c15caa3271.png)
divides but not , so it must be either or .
, then we have ![\[r\mid 17^ 2- 11^2 = 168\]](http://latex.artofproblemsolving.com/6/b/4/6b47534f7344a54c66d10a92c018e0cd77357dac.png)
Since is odd, 
, so 
. This implies , which we assumed to not be true.
, then , , as desired. 
, then ![\[3\cdot 2^{q-1} + 1 \mid 410\]](http://latex.artofproblemsolving.com/5/1/8/518ab9a03b7fcb764397a9476080a237af50ca97.png)
It's easy to check this is not possible., then ![\[3\cdot 7^{q-1} + 1 \mid 11^7 + 17^7\]](http://latex.artofproblemsolving.com/c/f/7/cf7f6509148b74a42448829368b5ce875d60b765.png)
Since 
, doesn't work. Now assume is odd. We have 
. So any prime factor of ![\[\frac{3\cdot 7^{q-1} + 1}{4}\]](http://latex.artofproblemsolving.com/0/6/c/06ce66aa1d00cb426cf99a55c7a5fe1073a8486e.png)
is 
, which implies 
, absurd. , then ![\[3^q + 1 \mid 11^3 + 17^3 =6244\]](http://latex.artofproblemsolving.com/f/a/6/fa67c5d4b75222db27fc3177e9212a0ae1f7c4ed.png)
One can manually check that this is not the case since 
..![\[3p+1 \mid 11^p + 17^p\]](http://latex.artofproblemsolving.com/3/6/6/3666c66a49a6254b5963ec0562e51877aaff885c.png)
Note that no prime that if divides , so the only prime factors of are and . However ![\[1\le \nu_2(3p+1) \le \nu_p(11^p + 17^p) = 2\]](http://latex.artofproblemsolving.com/7/4/e/74ec8613c7d559d3771b8ba6d707aa8a51665f5b.png)
and ![\[\nu_7(3p+1)\le \nu_7(11^p + 17^p) = 1\]](http://latex.artofproblemsolving.com/8/8/2/8824d1b589cdb59335623b3058589ecde7a7925e.png)
by LTE. So ![\[3p+1\in \{2,4,14,28\},\]](http://latex.artofproblemsolving.com/c/1/b/c1bc8b77f31f834bff5c5f3bf169713c66ace08c.png)
all of which don't work.
, so 
. Thus by Claim 15.1, ![\[3\cdot p^{q-1} + 1\equiv \{4,28\}\pmod{2p}\]](http://latex.artofproblemsolving.com/8/a/a/8aa211a563ff75f9c8946d48ce03cb68773f4208.png)
However, ![\[3\cdot p^{q-1} + 1\equiv p+1\pmod{2p},\]](http://latex.artofproblemsolving.com/f/0/5/f0516a896f537a598999582ecfe875735995db76.png)
so , contradiction.
It is easy to check 
so we will assume that 
satisfies 
then either 
or 
we have that 
If the order is 
we have that 
If the order is 
this works, and 
where is a product of primes that are 
We can easily see that 
by taking modulo 8.
we have that 
. We also know that by taking 
Thus, 
then 
By taking modulo , there is a contradiction.
then 
By taking modulo 
we have that 
is possible. However, this is impossible.
.
.
.
for example