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.
such that for all naturals 
the function satisfies:![\[a + b \mid a^{f(a)} + b^{f(b)} \]](http://latex.artofproblemsolving.com/e/b/b/ebb44041cedd29de516192d983a334d46e7fe7d6.png)
Bob wants to find all possible functions Alice could have. Help Bob and find all functions that Alice could have.
be the medial triangle of an acute-angle triangle 
. Suppose the line through 
perpendicular to 
meet 
at 
. Define 
analogously. Let 
. Similarly define 
and 
. Suppose the circle with diameter 
meet the -altitude at 
, where lies inside . Define 
and 
similarly. Let 
be the circumcenter of , and let 
be the circle with diameter 
, which meets 
at 
. Similarly define 
and 
.
are collinear.
lie on a circle centered at .
are coaxial. is perpendicular to the radical axis of .
denote the set of all real numbers. Find all functions 
such that
for all real numbers 
, 
.![\[x^2-cx+1 = \dfrac{f(x)}{g(x)},\]](http://latex.artofproblemsolving.com/1/3/3/1339acdaf1999e06e1a4b7dec7d1217389e2afef.png)
where 
and 
are polynomials with nonnegative real coefficients. For each 
, determine the minimum possible degree of , or show that no such 
exist.
such that every collection of consecutive positive integers contains an integer whose sum of digits is divisible by 
. for which he can do this, or prove that none exist.
, consider points on line such that 
are in that order and 
and 
. Similarly consider points 
on line 
, and 
on line . Prove that orthocenters of triangles 
and 
are equidistant from the circumcenter of .
be six positive real numbers. Prove that![\[a^2+b^2+c^2+\frac{4(ax+by+cz)\sqrt{ab+bc+ca}}{x+y+z} \geqslant 2(ab+bc+ca).\]](http://latex.artofproblemsolving.com/f/8/0/f808f7857c2150960f2a430837286487666797d8.png)
be a cyclic quadrilateral with circumcentre 
and circumcircle 
. Let 
be the intersection of tangents at 
and 
to . Let 
be the circumcircle of triangle 
and let 
, 
denote the second intersections of 
with respectively. Let 
and intersect at 
and 
be the circumcircle of triangle 
. If intersects again at 
, prove that the line tangent to at 
is also tangent to .
be an acute triangle with orthocenter 
. Let 
be the midpoint of , and 
be the intersection of the tangents from and to the circumcircle of . Denote by the circle centered at and tangent to line 
.
intersects at two distinct points , 
.
and 
meet at 
, while lines 
and 
meet at 
. Prove that either or lies on . +1 w
be a positive integer. Alice and Bob play the following game, with Alice going first after which they alternate turns. They determine the numbers 
in the following way.th turn, the player whose turn it is sets 
to be an integer such that:
There exists a polynomial with integer coefficients and 
for 
. In particular, she also loses if Bob is able to pick 
successfully. for which Alice can ensure that she wins regardless of Bob's strategy. be a scalene triangle with incircle . Denote by the midpoint of arc 
in the circumcircle of , and by 
the point where the -excircle touches . Suppose the circumcircle of 
meets again at 
and intersects at two points , .
or 
is tangent to .
and a point such that 
and 
are tangent to the circumcircle of . Let be the reflection of over and be the reflection of over . If is the circumcenter of 
, help Oscar prove that 
is perpendicular to .
denote the set of all positive integers. Find all real numbers 
for which there exists a function 
satisfying:
, the quantity 
is an integer if and only if 
;
, we have 
.
be a positive integer larger than 
, and let 
be integers. It is known that the equation 
has pairwise relatively prime integer roots. Prove that 
and 
are relatively prime.
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