AoPS Academy
be a sequence of real numbers, and let 
be a fixed positive integer less than 
. We say an index 
with 
is good if there exists some 
with 
such that 
, where the indices are taken modulo . Let 
be the set of all good indices. Prove that 
.
such that for any positive real numbers 
and 
,
, with 
.Initially, he folds the sheet along a straight line 
, where 
is a point on the side 
, so that vertex 
is located on side 
, as shown in the figure. Then folds the sheet again along a straight line 
, where 
is a point on side , so that vertex 
lies on the line ; and finally folds the sheet along the line 
. Matías observed that the vertices and 
were located on the same point of segment after making the folds. Calculate the measure of the angle 
.
are of the form 
, where 
, 2, 3, . For each , let 
be the greatest common divisor of 
and 
. Find the maximum value of as ranges through the positive integers.
for which 
is divisible by 
?
be a trapezium in which 
and 
. Let be then midpoint of the diagonal 
. If ![$[ABCD] = n \times [CDE]$](http://latex.artofproblemsolving.com/e/9/3/e9347ba2c8e9396adfe716159e711408a3b9dc69.png)
, what is the value of ?![$[t]$](http://latex.artofproblemsolving.com/8/a/8/8a86702dc60b9db916798b1115a6b6822bf828b9.png)
denotes the area of the geometrical figure
.)
be an odd prime and 
be integers so that the integers 
are divisible by . divides each of .
divides one of them, it divides all of them. Assume that it divides none of them. Then, there exists an inverse for all of them in modulo and then it's some algmanip to get 
a contradiction
divides one of them, it divides all of them. Assume that it divides none of them. Then, there exists an inverse for all of them in modulo and then it's some algmanip to get a contradiction
divides one of them, it divides all of them. Assume that it divides none of them. Then just do some things to get .
be an odd prime and be integers so that the integers . divides each of . divides any of we are done, so from now onwards we assume none of them are divisible by .Define 
such that
and 
.Then notice![\[\dfrac{1}{xy} = \dfrac{c}{a}\mod p\Longleftrightarrow (x\cdot y)^{2025}\equiv -1\mod p\]](http://latex.artofproblemsolving.com/e/b/6/eb6a6acc59325994d74209faa16c9408b6509521.png)
Claim : 
that is 
a contradiction to the fact that was an odd-prime.
, all of them are. So assume on the contrary that none of them are divisible by . So assume none of them are.
.
and 
. Multiplying these, we get,![\[ a^{2023 + 2025} \equiv b^{2023}c^{2025} \implies (a^2)^{2024} \equiv c^2 \cdot (bc)^{2023} \implies bc \cdot (bc)^{2023} \equiv c^2 \cdot (bc)^{2023} \implies b \equiv c \pmod{p}. \]](http://latex.artofproblemsolving.com/7/e/2/7e2a5811d0293d90d0803d797abe96dddb3cf6b6.png)
![\[ p\mid b^{2024} + c^{2024} \equiv 2b^{2024} \implies p \mid 2 \]](http://latex.artofproblemsolving.com/9/d/0/9d0ae48aed14e79a9bd3ac70d9e0de65b94a50a9.png)
which is a contradiction and we are done. 
divides all of divides one of , clearly divides all of , so assume divides none of them.
. The condition 
and we also have 
, so multiplying the three equations gives 
, so 
. This implies that 
, so 
, so 
. But 
, so dividing the two gives 
, so 
mod p. Since 
, we have 
. But then since 
, 
, so 
. Now, this means that 
is a multiple of , contradiction since doesn't divide 
and 
.
divides either of , then it divides everything and we're done. Henceforth, assume that 
. We will now work modulo with special powers of existence of inverses of .
, we get that 
. Swing this into 
to get 
. At last, putting this into 
, we get![\[a^{2025} + a^{2023}bc \equiv 0 \iff a^2 + bc \equiv 0 \pmod{p}.\]](http://latex.artofproblemsolving.com/9/f/3/9f3580a23b4e02a0cc044d1ed98e85e908c8dc66.png)
From here, we wish to eliminate completely modulo . This can be achieved via substituting 
. Putting this in we get![\[a^{2023} - \frac{a^{4046}}{c^{2023}} \equiv 0 \iff a^{2023} \equiv c^{2023} \pmod{p}.\]](http://latex.artofproblemsolving.com/9/7/6/976fafcba4a1cbc016f9da9b2282e872d622c737.png)
This yields![\[a^{2025} \equiv a^2c^{2023} \equiv -c^{2025} \pmod{p} \implies a^2 + c^2 \equiv 0 \pmod{p}.\]](http://latex.artofproblemsolving.com/a/f/e/afeed88cffe64ac63caa5078eb428e9dd101702e.png)
Finally, on combining this with 
, we get![\[p \mid c(c-b) \iff c \equiv b \pmod{p}\]](http://latex.artofproblemsolving.com/b/3/7/b3750c524d111659d1d05bb4c86672315b106d6a.png)
since 
. Upon putting 
in say , we get immediately get 
which is a contradiction since is odd and we're done. 
divides any one of them, it divides the other two as well. doesn't divide any of the three. Since is an odd prime, it has a primitive root, say 
.
.
where 
.
.
Cancelling and the denominator and taking mod 2 leads to a contradiction. Done!
be an odd prime and be integers so that the integers .?, but I suppose a lot of us got tunnel vision and didn't see the error because we already saw it in the test lol 
or 
or 
then divides all of them.
. is odd prime, i.e we would prove 
. we could take inverses.
we get
Multiplying equations 
, 
and 
we get
.
i.e
Now from we get
and from 1 we get
Let the order of 
mod be k.
i.e 
.
or
Putting this information in equation we get
or 
divides one of 
then it divides all of them, so henceforth assume it doesn't divide any of them. Thus, there exists a valid inverse in modulo for each of 
Then, we have 
and thus, multiplying all of them together, we have 
Thus, 
which gives 
Thus, 
which by putting in the first equation gives 
From the second equation, we get 
so 
so 
a contradiction.
for any 
and it follows for 
.
If divides any one of then 
.
W.L.O.G 
, now since 
and also 
. 
it suffices to disprove the fact that that 
,so FTSOC we assume that is the case.
does not divide any of 
. FTSOC 
, then 
, now since 
, it gives a contradiction. 
.
and 
and since 
similarly symmetrically other in and 
, but from 
we have 
, which implies 
we have 
and from 
we get this is equivalent to 
, but this is not possible as 
.. must divide .
, raise both sides to power 
to get terms 
and done
so all are divisible by 
be the order of 
mod p so
divides any one of , then it divides all the three numbers. Suppose 
. Then, 
but 
, which means that 
, forcing 
. Also, 
and 
which means 
, forcing 
., then and , which means , forcing . Since , we also have that from the earlier proof., then 
and 
, which means 
, forcing , which in turn forces . So, if divides any one of , then it divides all the three numbers.
. For integer 
and prime , we define the order of 
to be the smallest positive integer such that 
.
If is order of , 
and 
, then 
.
Suppose it was possible that 
. Note that if 
, then it will contradict the minimality of . So, 
. Hence, there exists 
and 
such that 
. Now, 
, which means 
, so 
. But since and 
, the minimality of gives us a contradiction. Hence, is impossible, and 
is forced. The proof is complete. ![\[\left(\frac{a}{b}\right)^{4046}\equiv1\pmod{p} \ \ \ \ (1)\]](http://latex.artofproblemsolving.com/0/3/c/03c40601bceb0fab5a6ffeeeba604aa68badcaad.png)
![\[\left(\frac{b}{c}\right)^{4048}\equiv1\pmod{p} \ \ \ \ (2)\]](http://latex.artofproblemsolving.com/9/4/9/949ad0a2ea48a5f0d3962fc499253bb2fbe1ca0b.png)
![\[\left(\frac{c}{a}\right)^{4050}\equiv1\pmod{p} \ \ \ \ (3)\]](http://latex.artofproblemsolving.com/5/9/c/59cc2f807076eebdc607ddf85468f6669909c174.png)
Multiplying these equations 
we obtain 
. So, 
.
is 
. So, ![\[\left(\frac{\left(bc\right)^{2023}}{b^{4046}}\right)\equiv\pm1\pmod{p},\]](http://latex.artofproblemsolving.com/5/9/8/598210f0185cbbeab376b8468ac11185c78c7702.png)
which gives ![\[\left(\frac{b}{c}\right)^{4046}\equiv1\pmod{p} \ \ \ \ (4)\]](http://latex.artofproblemsolving.com/9/7/5/975c0d72e9f1b8a5055adf740adbd3631ebc3120.png)
It follows that if is the order of 
, , then from equation , we get 
, and from equation 
, we get 
, using the 
in each case.
, so 
. Hence, we have that 
. So, 
.
, we get 
, and since , we have 
, which forces , as is an odd prime. However, if then , and are forced by our initial arguments.
divides any one of 
, then it divides all of them. Assume this to not be the case. Then we get the system of congruences 
Multiplying 
and 
, 
but from 
, 
Substituting in, 
Dividing from this new equation, 
Therefore 
a contradiction. 
Now assume that 
as if one of the three is 
modulo , all of them must be. Then we find that,
However then from this we may conclude that,
Dividing as 
modulo , we find that,
Now rewriting our conditions we have,
Squaring the first equation we have,
From the first equation we then find 
. Thus we have,
Thus we have from the original equations,
from which we easily see 
and hence are done.
, 
, 
.
From 
, we get that: 
, 
, 
, we have: 
, we get: 
and , we have: 
, 
, which is a contradiction! must divide one of . If divides any one of the numbers, must divide all the numbers individually, and we're done! 
=2023×2024×2025
And from this we get
and we are done
=2023×2024×2025And from this we getand we are done
, then![\[a^{2023}+b^{2023}\equiv 0 \mod p,\]](http://latex.artofproblemsolving.com/8/3/e/83eb190b5fa85d8dea225dcc29862a75566fb891.png)
![\[\iff b^{2023} \equiv 0 \mod p,\]](http://latex.artofproblemsolving.com/5/d/b/5dba54e9ecf89a47b19d5689c2329393b87d4596.png)
![\[\iff p\mid b,\]](http://latex.artofproblemsolving.com/a/7/b/a7bfece03e925f20b8a45a7d05193d2e9a90f21b.png)
and similarly we can find that . In general, using a similar proof, it can be shown that if 
, then divides each of , , and . FTSOC, assume that divides none of , , or .![\[\iff a^{2023}\equiv -b^{2023} \mod p,\]](http://latex.artofproblemsolving.com/8/1/8/818923601b1c3bd4663b2ffc84671be9a57925dc.png)
![\[\iff \left(\frac{a}{b}\right)^{2023}\equiv -1 \mod p,\]](http://latex.artofproblemsolving.com/e/6/8/e68ed1022f52ef7890eea80e85c20f4f99be4790.png)
![\[\iff ord_p\left(\frac{a}{b}\right) \mid 4046 \text{ and } 2\mid ord_p\left(\frac{a}{b}\right),\]](http://latex.artofproblemsolving.com/e/c/9/ec92d4eca021bc856f21fafb2c8557996a5c9459.png)
since we had that 
. Similarly, from the second equation, we get that![\[ord_p\left(\frac{b}{c}\right) \mid 4048 \text{ and } 16\mid ord_p\left(\frac{b}{c}\right).\]](http://latex.artofproblemsolving.com/a/8/9/a8920a410f2a1af044580f4fc60aa872c0ad23c1.png)
is prime, it is well-known that it must have a primitive root. Let said primitive root be and define and to be the unique integers such that 
and 
, where 
. Now, notice that if![\[2\mid ord_p\left(\frac{a}{b}\right) \text{ and } ord_p\left(\frac{a}{b}\right) \mid 4046,\]](http://latex.artofproblemsolving.com/5/8/8/588717e5e134f22fe7ffd8f045013a3ac830e228.png)
we get that![\[\nu_2\left(ord_p\left(\frac{a}{b}\right)\right)=1,\]](http://latex.artofproblemsolving.com/8/b/1/8b16c089c3c49f99785c489259e1351e37128591.png)
which in turn gives that![\[\nu_2(m)=\nu_2(p-1)-1,\]](http://latex.artofproblemsolving.com/b/4/9/b49456621127ee9da5266bbacdb544b1421ba7da.png)
since the order of mod is 
. Similarly, if![\[16\mid ord_p\left(\frac{b}{c}\right) \text{ and } ord_p\left(\frac{b}{c}\right) \mid 4048,\]](http://latex.artofproblemsolving.com/6/0/3/603243f29adb37cd296bbe937e2a3e675e5b7431.png)
then we have that![\[\nu_2(n)=\nu_2(p-1)-4.\]](http://latex.artofproblemsolving.com/c/2/0/c20759f7f4894678a21fbc57af30ccfb244b3a99.png)
![\[\frac{a}{c}\equiv \left(\frac{a}{b}\right)\left(\frac{b}{c}\right)\equiv g^{m+n} \mod p,\]](http://latex.artofproblemsolving.com/2/a/0/2a009851a1f21c5ce1a3d29247fce55cf2c20a78.png)
which means that![\[\nu_2\left(ord_p\left(\frac{a}{c}\right)\right)=\nu_2(p-1)-\nu_2(m+n).\]](http://latex.artofproblemsolving.com/1/c/a/1ca9ab1f825bd116747430272514ff490d684f41.png)
However, since![\[\nu_2(p-1)-1=\nu_2(m)>\nu_2(n)=\nu_2(p-1)-4,\]](http://latex.artofproblemsolving.com/0/f/8/0f8960478bd06900144fd543a3c64b96763a27e7.png)
we get that![\[\nu_2(m+n)=\nu_2(p-1)-4,\]](http://latex.artofproblemsolving.com/f/d/2/fd2ac92a256506f59ac0641161ef1c4fa8fb8f3e.png)
which gives that![\[\nu_2\left(ord_p\left(\frac{a}{c}\right)\right)=\nu_2(p-1)-(\nu_2(p-1)-4)=4.\]](http://latex.artofproblemsolving.com/3/3/1/331d4ab505bbc625d82d3d0318cfc588f3fbea58.png)
![\[c^{2025}+a^{2025}\equiv 0 \mod p,\]](http://latex.artofproblemsolving.com/b/a/d/bad74ec1ffd101b304144220701bf972c1d043f8.png)
![\[\iff a^{2025}\equiv -c^{2025} \mod p,\]](http://latex.artofproblemsolving.com/a/d/9/ad9f2857b261e693c9635e8ce64be7b7c8c88d1d.png)
![\[\iff \left(\frac{a}{c}\right)^{2025}\equiv -1 \mod p,\]](http://latex.artofproblemsolving.com/4/8/7/4878d5a9ad9e8600a3b35ef8ca89e05183c2aeda.png)
![\[\iff ord_p\left(\frac{a}{c}\right) \mid 4050,\]](http://latex.artofproblemsolving.com/6/b/d/6bd10eec121f089a055db571af01929381a55f99.png)
a contradiction, since 
does not divide 
. Therefore must divide each of , , and , as desired, finishing the problem.
divides one of 
, WLOG 
, we have 
, 
, so then divides all of them.
. We have the following modular congruences:![\[ \left(\frac{a}{b}\right)^{4046} \equiv 1 \pmod{p} \]](http://latex.artofproblemsolving.com/0/b/d/0bdd25f5d1ea41a0b381219a319f3677e2c899e9.png)
![\[ \left(\frac{b}{c}\right)^{4048} \equiv 1 \pmod{p} \]](http://latex.artofproblemsolving.com/4/2/d/42d5b135d3a6bff64503eebd1a8aec291a293688.png)
![\[ \left(\frac{c}{a}\right)^{4050} \equiv 1 \pmod{p} \]](http://latex.artofproblemsolving.com/5/0/2/50259c73c26a9290c222b1e1bb611237c452c9a3.png)
Firstly, observe that 
. Since 
, we have 
. Further, 
. From this, we conclude that:![\[ p \mid b^{2024}c + a^2b^{2023} = b^{2023}(bc + a^2) \implies a^2 \equiv -bc \pmod{p} \]](http://latex.artofproblemsolving.com/d/1/b/d1b561b2397fbdc7f8dca95b5ff62e8794e4bae3.png)
since 
.
, which, replacing in our first congruence, gives 
. Replacing this in our second congruence, we have![\[ \frac{b^{4046}}{c^{4046}} \cdot \frac{b^2}{c^2} \equiv 1 \pmod{p} \implies b^2 \equiv c^2 \pmod{p} \implies b^{2024} \equiv c^{2024} \pmod{p} \]](http://latex.artofproblemsolving.com/1/9/b/19b2e18df1dc97f85e66e923205936b9abc48abe.png)
Therefore, 
. Since is odd, we have 
, which gives 
as well, so we are done.
is divisible by would suffice. Assume all of them are invertible on 
. As 
we may say,
We then have 
. Since is invertible we are left with the key result,
Using the above fact, 
which deduces to 
and as 
we can easily show that 
. Now we are left with,
But we had already seen 
and is absurd, contradicting our assumption, hence none of them are invertible as desired.
![\[
\sum_{0 \leq k \leq l} (l - k) \binom{m}{k} \binom{q + k}{n}
= \binom{l + q + 1}{m + n + 1},
\]](http://latex.artofproblemsolving.com/b/5/8/b58c5b1f0f9c95266d189593f02bae85c100c82a.png)
![\[
\text{integers } l, m \geq 0,\quad \text{integers } n \geq q \geq 0.
\]](http://latex.artofproblemsolving.com/8/e/2/8e25f8d7e35ac7813e40af7cc4806000e70999af.png)
and 
Prove that![$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$](http://latex.artofproblemsolving.com/5/6/1/561a990e5885863c77d454ffa3f1973e1bbf60b4.png)
![$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$](http://latex.artofproblemsolving.com/4/b/1/4b14911672688250c909571d788aaf10724462eb.png)
Where 
)
Prove that
and 
Note that we can choose 
such that, 
cause of periodicity of modulo 
.
such that 
as, 
,
then 
divides 
,
divides 
and 
divides 
such that, 
Then by 2nd equation given in the question that is 
we have 
but by 
we have 
which is a contradiction as 
But, due to the order of an integer modulo 
a multiple of 
,
.
,
and 
.
,
.
.
,
.