It's February and we'd love to help you find the right course plan!

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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

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0 replies
jlacosta
Feb 2, 2025
0 replies
P6 Geo Finale
math_comb01   7
N 2 minutes ago by GuvercinciHoca
Source: XOOK 2025/6
Let $ABC$ be a triangle with incenter $I$ and excenters $I_A$, $I_B$, $I_C$ opposite to $A,B,C$ respectively. Suppose $BC$ meets the circumcircle of $I_AI_BI_C$ at points $D$ and $E$. $X$ and $Y$ lie on the incircle of $\triangle ABC$ so that $DX$ and $EY$ are tangents to the incircle (different from $BC$). Prove that the circumcircles of $\triangle AXY$ and $\triangle ABC$ are tangent.

Proposed by Anmol Tiwari
7 replies
math_comb01
Feb 10, 2025
GuvercinciHoca
2 minutes ago
A functional equation
super1978   1
N 13 minutes ago by pco
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(y-x)-xf(y))+f(x)=y(1-f(x)) $$for all $x,y \in \mathbb R$
1 reply
super1978
an hour ago
pco
13 minutes ago
Sequences Handout
M11100111001Y1R   4
N 17 minutes ago by MR.1
Source: Own
Hi everyone, I wrote this handout about sequences in NT.
Hope you enjoy!
4 replies
+2 w
M11100111001Y1R
Oct 19, 2022
MR.1
17 minutes ago
[Handout] 50 non-traditional functional equations
gghx   2
N 44 minutes ago by GreekIdiot
Sup guys,

I'm retired. I love FEs. So here's 50 of them. Yea...

Functional equations have been one of the least enjoyed topics of math olympiads in recent times, mostly because so many techniques have been developed to just bulldoze through them. These chosen problems do not fall in that category - they require some combi-flavoured creativity to solve (to varying degrees).

For this reason, this handout is aimed at more advanced problem solvers who are bored of traditional FEs and are up for a little challenge!

In some sense, this is dedicated to the "covid FE community" on AoPS who got me addicted to FEs, people like EmilXM, hyay, IndoMathXdZ, Functional_equation, GorgonMathDota, BlazingMuddy, dangerousliri, Mr.C, TLP.39, among many others: thanks guys :). Lastly, thank you to rama1728 for suggestions and proofreading.

Anyways...
2 replies
gghx
Sep 23, 2023
GreekIdiot
44 minutes ago
No more topics!
Strange NT
magicarrow   19
N Nov 30, 2023 by asdf334
Source: Romanian Masters in Mathematics 2020, Problem 6
For each integer $n \geq 2$, let $F(n)$ denote the greatest prime factor of $n$. A strange pair is a pair of distinct primes $p$ and $q$ such that there is no integer $n \geq 2$ for which $F(n)F(n+1)=pq$.

Prove that there exist infinitely many strange pairs.
19 replies
magicarrow
Mar 1, 2020
asdf334
Nov 30, 2023
Strange NT
G H J
G H BBookmark kLocked kLocked NReply
Source: Romanian Masters in Mathematics 2020, Problem 6
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magicarrow
146 posts
#1 • 11 Y
Y by DPS, pinetree1, amar_04, Loppukilpailija, NJOY, Math-wiz, Idio-logy, MathbugAOPS, centslordm, MatBoy-123, megarnie
For each integer $n \geq 2$, let $F(n)$ denote the greatest prime factor of $n$. A strange pair is a pair of distinct primes $p$ and $q$ such that there is no integer $n \geq 2$ for which $F(n)F(n+1)=pq$.

Prove that there exist infinitely many strange pairs.
This post has been edited 1 time. Last edited by magicarrow, Mar 1, 2020, 11:19 AM
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Loppukilpailija
155 posts
#2 • 19 Y
Y by Dem0nfang, p_square, amar_04, NJOY, fjm30, Supercali, leibnitz, Math-wiz, bumjoooon, Polynom_Efendi, mijail, Mathhunter1, centslordm, ILOVEMYFAMILY, Kobayashi, Kingsbane2139, LoloChen, pavel kozlov, Mango247
Solution.

Remark 1.

Remark 2.
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Dem0nfang
478 posts
#3 • 1 Y
Y by Mango247
Well, what I will say might seem weird, but $n$ and $n+1$ are co-prime, so their largest prime factors will be different.

Doesn't this imply that $\exists$ infinitely many 'strange' pairs?

@2below What is Kobayashi?
This post has been edited 3 times. Last edited by Dem0nfang, Mar 1, 2020, 10:56 AM
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Sugiyem
115 posts
#4 • 2 Y
Y by trololo23, Mango247
I only found 304 pairs during contest tho :(
This post has been edited 1 time. Last edited by Sugiyem, Mar 1, 2020, 12:16 PM
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ubermensch
820 posts
#5
Y by
Perhaps(rather probably) I’m wildly misinterpreting the problem, but why shouldn’t $n=p^{\alpha}$ with Kobayashi’s finish this off?
What am I saying, just choosing $n=p$ kills this... there’s probably a typo in the problem...
This post has been edited 1 time. Last edited by ubermensch, Mar 1, 2020, 10:45 AM
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YII.I.
94 posts
#6
Y by
I don't understand the question. Since $(n,n+1)=1$, $F(n)F(n+1)$ must be a product of two distinct primes. So trivial?
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_el_doopa
62 posts
#7 • 1 Y
Y by centslordm
The problem statement is obviously wrong. I think that you need to proof that the complement of strange pairs is infinite. Or the definition should be adjusted.
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WypHxr
189 posts
#8
Y by
Are you sure your statement is right?
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p_square
442 posts
#9 • 4 Y
Y by YII.I., Rg230403, amar_04, aops29
actual problem wrote:
For each integer $n \geq 2$, let $F(n)$ denote the greatest prime factor of $n$. A strange pair is a pair of distinct primes $p$ and $q$ such that there is $\textbf{no}$ integer $n \geq 2$ for which $F(n)F(n+1)=pq$.

Prove that there exist infinitely many strange pairs.
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magicarrow
146 posts
#10
Y by
Corrected problem statement; thanks @above
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heron1000
138 posts
#13
Y by
https://mathoverflow.net/questions/332901/greatest-prime-factor-of-n-and-n1 has a bit more on this problem.
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rmtf1111
698 posts
#14
Y by
Let $p$ and $q$ be primes such that $q \mid 2^p-1$ and $q\neq F(2^p-1).$ Clearly, there are infinitely many such primes $q.$ Suppose there exists an integer $m$ such that $F(m)F(m+1)=2q,$ then $m$ must be equal to $2^u-1$ for some $u.$ Now, $q$ divides $2^u-1$ iff $p$ divides $u,$ which would imply that $F(m)\neq q,$ hence we achieve a contradiction.


EDIT: Whoops, this is wrong, we do not know if there are infinitely many composite numbers of the form $2^p-1$ for $p$ prime, see #2.
This post has been edited 2 times. Last edited by rmtf1111, Mar 3, 2020, 4:15 PM
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IndoMathXdZ
691 posts
#15 • 3 Y
Y by Sugiyem, Awesome_guy, Aryan-23
Should have spent more effort on problem 6 rather than problem 5 in the contest. Huge mistake :(
Here's a solution with motivation:
The condition of the problem somehow forces us to take either $p$ or $q$ being $2$. Otherwise, it is really hard to deal with something like $2^a \cdot 3^b \pm 1$.
We will prove that the sufficient condition $o_p(2) = o_q(2)$ for two distinct primes $p > q$ implies that $(2,q)$ is strange.
Claim 01. If $p$ and $q$ are distinct odd primes such that $o_p (2) = o_q (2)$ and $p > q$, then $(2,q)$ is strange.
Proof. Notice that since either $F(n)$ or $F(n+1)$ is $2$, then we would have $n = 2^k$ or $n = 2^k - 1$. We would like to prove that for any $q$ such that $q | 2^k - 1$, then $F(2^k - 1) \not= q$, and for any $q$ such that $q | 2^k + 1$, we would have $F(2^k + 1) \not= q$.
If $o_p (2) = o_q (2)$, then we would have $p | 2^k - 1$ as well if $q | 2^k - 1$. Hence, $F(2^k - 1) \ge p > q$.
Furthermore, if $q | 2^k + 1$, then $q | 2^{2k} - 1$. Suppose $p | 2^k - 1$, then we would have $q | 2^k - 1$, as well, forcing $q | 2$, which is a contradiction. Hence, $p | 2^k + 1$ as well, which forces $F(2^k + 1) \ge p > q$.
It suffices to prove that there are infinitely many distinct odd primes $p,q$ such that $o_p (2) = o_q (2)$. This is equivalent to finding infinitely many numbers $i$ such that $2^i - 1$ (or $2^i + 1$) have two distinct primes that doesn't appear in any $2^j - 1$, $j < i$ ($2^j + 1$).
Now, dealing with order and stuffs motivates us to consider $i$ having as least divisors as possible.
Playing with $2^i + 1$ has an advantage instead of $2^i - 1$ because it restricts the order of the primes, which is why I consider $2^i + 1$. Playing with $2^p + 1$ isn't helpful, as we can't factorized it (since we want something such that $2^i + 1$ produces at least 2 new prime, so factoring is a nice option.)
Now, consider $2^{2p} + 1$. Since $p$ is odd. Then, by Sophie Germain Identity, we have
\[ 2^{2p} + 1 = (2^p + 2^{\frac{p + 1}{2}} + 1)(2^p - 2^{\frac{p + 1}{2} } + 1) \]Now, consider the following lemma.
$\textbf{Lemma 02.}$ If an odd prime $q$ satisfy $q | 2^{2p} + 1$, and $q \not= 5$, then $o_q (2) = 4p$.
Proof. Notice that $q | 2^{4p} - 1$. Hence, $o_q (2) | 4p$. But if $o_q (2) | 2p$, then $q | 2^{2p} - 1$, which forces $q | 2$, a contradiction.
Hence, we would have $o_q (2) \in \{ 4, 4p \}$. But if $o_q (2) = 4$, then we would have $q | 2^2 + 1 = 5$. But we assume that $q \not= 5$, hence we would have $o_q (2) = 4p$.
Now, it's tempting to consider $p$ modulo $20$ since we would like to see the behavior of $2^{2p} + 1$ in modulo $25$. Now, notice that if $p \equiv 1 \ (\text{mod} \ 20)$, then we would have $2^{\frac{p + 1}{2}} \equiv 2, -2 \ (\text{mod} \ 25)$.
Hence, one of $2^p + 2^{\frac{p + 1}{2}} + 1$ or $2^p - 2^{\frac{p + 1}{2}} + 1$ is $5$ modulo $25$, and the other is $1$ modulo $25$.
Confirming that $2^p + 2^{\frac{p + 1}{2}} + 1 > 2^p - 2^{\frac{p + 1}{2}} + 1 > 5$ for any prime $p \equiv 1 \ (\text{mod} \ 20)$, and notice that both factors can't have the same odd prime divisor, since it would divide $2 \cdot 2^{\frac{p + 1}{2}}$. Then, each of these factors contribute a new prime $q$ with order $4p$. Therefore, we would have two distinct primes $q_1$ and $q_2$ such that
\[ o_{q_1} (2) = o_{q_2} (2) = 4p \]We are hence finished.

Remark. In the contest, my friend found a solution using $(2, G(2^{2^n} - 1))$ where $G(n)$ is the smallest prime factor of $n$, and $2^{2^n} - 1$ is composite. But unfortunately, the number of composite Fermat number remains open till now.
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PedroBigMan
14 posts
#17
Y by
rmtf1111 wrote:
Let $p$ and $q$ be primes such that $q \mid 2^p-1$ and $q\neq F(2^p-1).$ Clearly, there are infinitely many such primes $q.$ Suppose there exists an integer $m$ such that $F(m)F(m+1)=2q,$ then $m$ must be equal to $2^u-1$ for some $u.$ Now, $q$ divides $2^u-1$ iff $p$ divides $u,$ which would imply that $F(m)\neq q,$ hence we achieve a contradiction.


EDIT: Whoops, this is wrong, we do not know if there are infinitely many composite numbers of the form $2^p-1$ for $p$ prime, see #2.

Am i being dumb or doesnt Catalan Conjecture say this?
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i3435
1348 posts
#18
Y by
Catalan's conjecture says that no perfect powers are 1 apart, except for $8$ and $9$.
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TheUltimate123
1739 posts
#19
Y by
Solved with Th3Numb3rThr33.

We will show there are infinitely many strange pairs of the form \((2,q)\).

Claim: For primes \(q<r\), if \(\operatorname{ord}_q(2)=\operatorname{ord}_r(2)\) then \((2,q)\) is a strange pair.

Proof. First note that for \(F(n)F(n+1)\) to be even, there must be a power of two among \(n\), \(n+1\), so \(n\) is either of the form \(2^k-1\) or \(2^k\). Therefore it suffices to show whenever \(q\mid2^k\pm1\) that \(F(2^k\pm1)\ne q\).

Let \(d\) be the common order. Then:
  • If \(q\mid2^k-1\), observe from \(2^d-1\mid2^k-1\) that \(F(2^k-1)\ge F(2^d-1)\ge r>q\).
  • Suppose \(q\mid2^k+1\). Observe that for general primes \(p\), we have \(p\mid2^k+1\) if and only if \(\operatorname{ord}_p(2)\) is even and \(k\equiv\frac12\operatorname{ord}_p(2)\pmod{\operatorname{ord}_p(2)}\).
    Thus \(q\mid 2^k+1\iff k\equiv\frac12d\pmod d\iff r\mid2^k+1\), so \(F(2^k+1)\ge r>q\).
Thus \((2,q)\) is strange. \(\blacksquare\)

Select a large prime \(p\). First I claim we can find prime divisors \(5<q<r\) of \(2^{2p}+1\). Note that \(2^{2p}+1\) is not a multiple of three and \[\nu_5\left(2^{2p}+1\right)=\nu_5\left(4^p-(-1)^p\right)=1.\]By Sophie-Germain we have \[2^{2p}+1=\left(2^p-2^{(p+1)/2}+1\right)\left(2^p+2^{(p+1)/2}+1\right).\]These two factors are relatively prime. One of these factors is divisible by 5, so divide 5 out. Then what remains are two relatively prime divisors each with prime factors greater than 5, so \(q\) and \(r\) exist.

Now note that \(\operatorname{ord}_q(2)=\operatorname{ord}_r(2)=4p\) (since neither order equals 4). We have exhibited infinitely many \(p\), and therefore infinitely many \(q\). This completes the proof.
This post has been edited 1 time. Last edited by TheUltimate123, Dec 24, 2020, 8:22 AM
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GeronimoStilton
1521 posts
#20 • 1 Y
Y by Mango247
Solution with awang11.

We generate infinitely many primes $q$ for which $F(n)F(n+1)=2q$ has no solution.

Let $p$ be a particular odd prime congruent to $3$ modulo $8$ (of which there are infinitely many by Dirichlet's Theorem on Arithmetic Progressions). Observe the factorization
\[2^{2p}+1=(2^p-2^{(p+1)/2}+1)(2^p+2^{(p+1)/2}+1).\]Moreover, note the two factors are relatively prime, as they have difference $2^{(p+3)/2}$ which certainly divides neither. Since $p\equiv 3\pmod{8}$, we have $5\mid 2^p-2^{(p+1)/2}+1$ and $5\nmid 2^p+2^{(p+1)/2}+1$. Additionally, $5$ divides the expression exactly once by LTE. So let $q$ be the minimal prime divisor of $2^{2p}+1$ that is not $5$. Clearly $q$ is not also the largest prime factor of $2^{2p}+1$.

Claim: $q\mid 2^k+1$ iff $2p\mid k,4p\nmid k$ and $q\mid 2^k-1$ iff $4p\mid k$.

Solution: For the latter case, apply the Euclidean algorithm to see we get $q\mid 2^{\gcd(k,4p)}-1$. Clearly $\gcd(k,4p)=1,2,4$ cannot occur and neither can $\gcd(k,4p)=p,2p$ by assumption, so $4p\mid k$. For the former case, note $q\mid 2^{2k}-1$ so by the second part of the claim this implies $4p\mid 2k$, and clearly $4p\nmid k$. $\fbox{}$

Now, we argue that $F(n)F(n+1)=2q$ cannot occur. If $n=2^k$ for some $k$ then the claim suffices by showing $F(n+1)\ge F(2^{2p}+1)>q$ and similarly if $n=2^k-1$ for some $k$ then $F(n)\ge F(2^{2p}+1)>q$, since either way we get $q$ divides the $2^k\pm1$ term or we are automatically done. The primes $q$ generated are clearly distinct for each such prime $p$ because of the claim, so we are done.
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squareman
966 posts
#21 • 3 Y
Y by rama1728, Catsaway, centslordm
Solved with rama1728.

Claim: If for two odd primes $p < q$ we have $\operatorname{ord}_q(2) = \operatorname{ord}_p(2)$ then $(p,2)$ is strange.

Proof: If $F(n)F(n+1)=2p,$ then either $(n,n+1)$ is either $(2^k, 2^k+1)$ or $(2^k-1, 2^k)$ for some positive integer $k.$ In the latter case, note $p \mid 2^{k}-1 \implies \operatorname{ord}_p(2) \mid k \implies \operatorname{ord}_q(2) \mid k \implies q \mid 2^k-1.$ In the former case, $p \mid 2^{k}+1 \implies \operatorname{ord}_p(2) =\operatorname{ord}_q(2) = 2k \implies q \mid 2^{2k}-1,$ and since $q \nmid 2^k-1$ because $\operatorname{ord}_q(2) \ne k,$ we have $q \mid 2^k+1.$ $\square$

It suffices to show there are infinitely many distinct pairs of odd primes $p,q$ where $\operatorname{ord}_q(2) =  \operatorname{ord}_p(2).$ Take any prime $P = 2k+1 > 5$ (obviously infinitely many) and note
\begin{align*}
2^{2P} + 1 &=4\cdot 2^{4k} + 1\\
&=(2^{2k+1}+2^{k+1}+1)(2^{2k+1}-2^{k+1}+1).
\end{align*}Note $\gcd(2^{2k+1}+2^{k+1}+1, 2^{2k+1}-2^{k+1}+1)=\gcd(2^{k+2}, 2^{2k+1}-2^{k+1}+1)=1,$ $3 \nmid 2^{2P}+1,$ and $25 \nmid 2^{2P}+1$ since $10 \nmid 2P.$ So we can take two distinct primes $p,q > 5$ dividing $2^{2P}+1,$ so $\operatorname{ord}_q(2), \operatorname{ord}_p(2) \mid 4P.$ Obviously now $\operatorname{ord}_q(2), \operatorname{ord}_p(2) > 4,$ so $\operatorname{ord}_q(2) = \operatorname{ord}_p(2) = 4P$ as desired. $\blacksquare$
This post has been edited 2 times. Last edited by squareman, Jan 12, 2022, 2:25 AM
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IAmTheHazard
4998 posts
#22 • 1 Y
Y by centslordm
stupid silly problem. got all the right ideas until i didnt remember SOPHIE GERMAIN so i am dead lmao fascinating

Consider numbers of the form $2^{2p}+1$ for $p$ an odd prime. Note that this number is always divisible by $5$ and never by $3$.

Claim: For sufficiently large $p$, $2^{2p}+1$ has at least two prime divisors other than $5$.
Proof: We use SOPHIE GERMAIN:
$$2^{2p}+1=4\cdot (2^{\frac{p-1}{2}})^4+1^4=(2^p+2^{\frac{p+1}{2}}+1)(2^p-2^{\frac{p+1}{2}}+1).$$Both of these factors are integers, which are coprime (since their gcd divides $2^{\frac{p+3}{2}}$ by using the Euclidean algorithm) and larger than $5$ for all large enough $p$, which implies the desired result. $\blacksquare$

Now, in general, if $p \neq q$ are odd primes, then
$$\gcd(2^{2p}+1,2^{2q}+1) \mid \gcd(2^{4p}-1,2^{4q}-1)=2^{\gcd(4p,4q)}-1=2^4-1=15,$$which implies that it is $5$. Therefore, we can find infinitely many primes $p>5$ such that $p$ divides $2^{2q}+1$ for some (unique) $q$ but $F(2^{2q}+1) \neq p$ by taking the smaller of the two prime divisors supplied by the above claim across all large $q$. I claim that, for these $p$, $(2,p)$ is always a strange pair.

First note that $\mathrm{ord}_p(2)$ is either $4$ or $4q$, but the former would imply that $p \in \{3,5\}$, so it must equal $4q$. Now, if $F(n)F(n+1)=2p$ for some $n$, then either $n$ or $n+1$ is a power of $2$. In the first case, we would have $F(2^k+1)=p$ for some $k$. It is necessary (and sufficient) to have $k \equiv 2q \pmod{4q}$, but this implies that the other prime $p'>p$ dividing $2^{2q}+1$ also divides $2^k+1$: contradiction. In the second, we have $F(2^k-1)=p$ for some $k$. It is necessary (and sufficient) to have $k \equiv 0 \pmod{4q}$, but this again implies that $p'$ (as defined before) also divides $2^k-1$. This implies the desired result. $\blacksquare$
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asdf334
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Well that was interesting. like @above, I got all the right ideas but focused on $2^p-1$ for prime $p$. I guess I didn't look at the big picture and realize that I needed an explicit factorization trick (new hard skill in the toolkit!)
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