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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Inequality with a,b,c
GeoMorocco   6
N 10 minutes ago by GeoMorocco
Source: Morocco Training
Let $   a,b,c   $ be positive real numbers such that : $   ab+bc+ca=3   $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
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GeoMorocco
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GeoMorocco
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Centroid, altitudes and medians, and concyclic points
BR1F1SZ   1
N 10 minutes ago by sami1618
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
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BR1F1SZ
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KhuongTrang   31
N 15 minutes ago by NguyenVanHoa29
Source: own
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anirbanbz
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Today at 4:14 PM
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sqing   11
N Today at 3:02 PM by sqing
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sqing
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giangtruong13   4
N Today at 1:26 PM by sqing
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N Today at 1:19 PM by Jackson0423
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lifeismathematics   2
N Today at 12:09 PM by Adywastaken
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lifeismathematics
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Adywastaken
Today at 12:09 PM
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sqing   7
N Today at 10:31 AM by sqing
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
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$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
7 replies
sqing
Yesterday at 12:46 PM
sqing
Today at 10:31 AM
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
Valentin Vornicu   66
N Apr 25, 2025 by Ilikeminecraft
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
66 replies
Valentin Vornicu
Oct 24, 2005
Ilikeminecraft
Apr 25, 2025
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
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joshualiu315
2534 posts
#55
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The answer is $\boxed{\text{yes}}$.

Let $p_x$ be a prime that divides $2^{3^x}+1$ but not $2^{3^{i}}+1$ for $i<x$. This prime will always exist by Zsigmondy, meaning our desired number is

\[n=3^{2000}p_2p_3\cdots p_{2000}.\]
Seeing that this works is trivial, so we are done.
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shendrew7
794 posts
#56
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We note $n_1=3^2$ satisfies the condition. We aim to find a prime $p$ such that $pn_i$ satisfies the condition and $\gcd(p,n_i)=1$, so it suffices to have
\[2^{pn} \equiv -1 \pmod{p} \implies 2^{2n} \equiv 1, 2^n \not\equiv 1 \pmod{p}.\]
This primitive root of 2 modulo $p$ exists by Zsigmondy, as desired. Hence we can induct to find $n_{2000}$ with the desired 2000 prime divisors. $\blacksquare$
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peppapig_
281 posts
#57
Y by
Overkill?

In general, I claim that there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$.

I prove this by claiming that if there exists a positive integer $n$ with $k$ distinct integer divisors in the form of
\[n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k},\]such that $p_1<p_2\dots<p_k$, then there exists a prime $p_{k+1}>p_k$ such that for all positive integers $e_{k+1}$, it is true that $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the problem condition, where we can make $m$ any positive integer we want.

C1: First, before we begin, I claim that if any prime $q\mid n$, then $nq$ also satisfies problem conditions. This will then prove the part where we can make $m$ whatever we want. This is because
\[n\mid 2^n+1 \iff \nu_q(n)\leq \nu_q(2^n+1),\]and by LTE (which we can use, since $n\mid 2^n+1$ implies that $q\mid 2^n+1$), this gives us that
\[\nu_q(nq)=1+\nu_q(n)\leq 1+\nu_q(2^n+1)=\nu_q(2^{nq}+1),\]which means that $nq$ also satisfies the problem conditions, as desired. Therefore, by induction, we also get that $nq^m$ also satisfies problem conditions.

C2: Now, I claim that there exists a prime $p_{k+1}$ such that for all positive integers $e_{k+1}$, $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the conditions. For simplicity, from here on out, I will refer to $p_{k+1}$ as $r$ and $p_k$ as $q$. Note that this implies that
\[r\mid 2^n+1 \iff ord_r(2)\mid 2n,\]and since $ord_r(2)\mid \phi(r)=r-1$, we get that $ord_r(2)\mid \gcd(2n,r-1)$. First, I claim that there always exists an $r$ that divides $2^{q^c}+1$ for some $c$ we can choose such that $r>q$. This is clearly true by Zsigmondy's theorem. I now claim that if we take this $r$, it is true that $rn$ also satisfies problem conditions. This, combined with (C1), will prove our master claim, which states that $nr^e_{k+1}$ satisfies problem conditions for any $e_{k+1}$.

To prove this, first make sure that our previous $n$ is divisible by the $q^c$ we chose. We can do this by making another $n$ that satisfies problem conditions by multiplying it by a power of $q$, which we can do by (C1). Next, by LTE, we have,
\[\nu_r(rn)=1\leq \nu_r(2^{rn}+1)=\nu_r(2^n+1)+1,\]which we can do since we know that $q^c \mid n$ and $r\mid 2^{q^c}+1$, which gives that $r\mid 2^n+1$. This means that $rn$ satisfies problem conditions, since $rn\mid 2^{rn}+1$, which combined with (C1), proves our inductive step claim.

Finally, to complete our induction, note that $3\mid 2^3+1$. Therefore, there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$, finishing the problem.
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SenorSloth
37 posts
#58
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We claim that the answer is yes. We will induct to show that if there exists such an $n$ with $x$ distinct prime factors, then there also exists such an $n$ with $x+1$ distinct prime factors. By repeating this logic we can find an $n$ with exactly $2000$ distinct prime factors.

Our base case is $n=9$, which works since $9\mid 513$. By Zsigmondy, for any $n>3$, there exists a prime $p$ dividing $2^n+1$ that does not divide any $2^k+1$ for smaller $k$. Since this implies that $2$ has order $2n$ modulo $p$, the prime is at least $2n+1$ and thus cannot divide $n$. Thus we know that $pn\mid 2^n+1$ and consequently $pn \mid 2^{pn}+1$. $pn$ has $1$ more distinct prime factor than $n$, so the induction is complete.
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Niku
120 posts
#59
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Do you realise that this post was made 20 years ago.
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BestAOPS
707 posts
#60
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Overkill proof while forgetting about Zsigmondy:

Define two sequences as follows: $n_0 = 1$; $p_i$ is (a) the smallest prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$, or (b) if that doesn't exist, the smallest prime factor of $2^{n_i} + 1$; and $n_{i+1} = n_ip_i.$

Notice that each $n_i$ is divisible by all of the previous ones, and that all $n_i$ and $p_i$ are odd.

First, we show that all $n_i$ satisfy $n_i \mid 2^{n_i} + 1$. We proceed by induction. We can see that $n_0 = 1$ works, so assume $n_i$ works (and all the ones before it). We want to prove $n_{i+1} \mid 2^{n_{i+1}} + 1$.

Note that if a prime $p$ divides $n_{i+1}$, then $p = p_j$ for some $j$ satisfying $0 \leq j \leq i$.
This also means that $p_j$ is a factor of $2^{n_j} + 1$.
Then, by LTE, we have
\[ \nu_{p_j}(2^{n_{i+1}} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}\left(\frac{n_{i+1}}{n_j}\right) = \nu_{p_j}(n_{i + 1}) + \nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j). \]However, the strong inductive hypothesis implies $\nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j) \geq 0$, so we have $\nu_{p_j}(2^{n_{i+1}} + 1) \geq \nu_{p_j}(n_{i+1})$.
As this is true for all $p$, the inductive step is complete.

Next, we claim that eventually, the number of distinct prime factors of $n_i$ is always one more than the number of distinct prime factors of $n_{i-1}$. This is equivalent to showing that eventually, there always exists a prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$.

Suppose, for some $i$, that every prime factor of $2^{n_i} + 1$ is also a factor of $n_i$.
Then, let $p$ be a prime factor of $n_i$, and pick the minimal $j$ such that $p_j = p$. This minimality implies that $\nu_{p_j}(n_j) = 0$.
We have
\[ \nu_{p_j}(2^{n_i} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i) - \nu_{p_j}(n_j) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i). \]We can raise $p_j$ to the power of both sides to get
\[ {p_j}^{\nu_{p_j}(2^{n_i} + 1)} = {p_j}^{\nu_{p_j}(2^{n_j} + 1)} {p_j}^{\nu_{p_j}(n_i)}. \]Doing this for every prime factor $p$ of $n_i$ (notice that $j$ is now a one-to-one function of $p$) and multiplying the resulting equations, we get
\[ 2^{n_i} + 1 = n_i \prod_p p^{\nu_p(2^{n_{j(p)}} + 1)}. \]For all $p$, we have $p^{\nu_p(2^{n_{j(p)}} + 1)} \leq 2^{n_{j(p)}} + 1$. Thus,
\[ 2^{n_i} + 1 \leq n_i \prod_p (2^{n_{j(p)}} + 1). \]Since $j$ is one-to-one, every $j(p)$ is unique and in the set $\{0, 1, \ldots, i-1\}$. Therefore, we have the inequality
\[ 2^{n_i} + 1 \leq n_i \prod_{j=0}^{i-1} (2^{n_j} + 1). \]Taking the log base $2$ of both sides, we have
\[ n_i < \log _2 (2^{n_i} + 1) \leq \log _2 n_i + \sum _{j=0}^{i-1} \log_2(2^{n_j} + 1) < \log_2 n_i + \sum _{j=0}^{i-1} (n_j + 1) = \log_2 n_i + i + \sum_{j=0}^{i-1}n_j. \]Now, in order to achieve a bound on $n_i$, we notice that $p_i \geq 3$ for all $i$, so therefore, $n_i \geq 3^i$. It is then easy to see that $\sum_{j=0}^{i-1}n_j \leq \frac12 n_i$ for all $i \geq 1$. Then,
\[ n_i < \log_2 n_i + i + \frac12 n_i \leq \log_2 n_i + \log_3 n_i + \frac12 n_i \iff n_i < 2(\log _2 n_i + \log _3 n_i). \]This inequality obviously cannot be satisfied as $n_i$ grows large. Thus, eventually, we must have that there exists a prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$. Hence, there must eventually exist an $n$ in our sequence with exactly 2000 distinct prime factors.
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LQFNB
12 posts
#61
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是否存在一个恰有2000个素因子的正整$n$$n \mid 2^n + 1$?
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LQFNB
12 posts
#62
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是否存在一个恰有2000个不同素因子的正整$n$$n \mid 2^n + 1$
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eibc
600 posts
#63
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The answer is yes. We will construct a positive integer $n = p_1^2 p_2p_3 \cdots p_{2000}$ where $p_1 < p_2 < \cdots < p_{2000}$ are distinct primes such that $n \mid 2^n + 1$.

First, we set $p_1 = 3$. For $i \ge 2$, let $q_i = 3^2 \cdot p_2 \cdots p_3 \cdots p_{i - 1}$. Then, we shall construct $p_i$ recursively by picking a primitive divisor of $2^{2q_i} - 1$, which exists by Zsigmondy's theorem.

Now, we show that $n \mid 2^n + 1$. It suffices to show that for $1 \le i \le 2000$, we have $\nu_{p_i}(2^n + 1) \ge \nu_{p_i}(n)$. For $i = 1$, we note that by LTE,
$$\nu_3(2^n + 1) = \nu_3(2 + 1) + \nu_3(n) = 1 + \nu_3(n) > 1.$$For $i > 1$, we note that by construction, $\text{ord}_{p_i}(2) = 2q_i$. Thus, $2^{q_i} \equiv -1 \pmod {p_i}$, so by LTE, we have
$$\nu_{p_i}(2^n + 1) = \nu_{p_i}((2^{q_i})^{n/q_i} + 1) = \nu_{p_i}(2^{q_i} + 1) + \nu_{p_i}(n/q_i) \ge 1 + \nu_{p_i}(n) > \nu_{p_i}(n),$$so we are done.
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Natrium
50 posts
#64
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Lemma. If $a\mid 2^a+1$ and $b\mid 2^b+1$, where $a=3^\alpha a'$, $b=3^\beta b'$, $3 \nmid a', b'$ and $a'$ and $b'$ are coprime, then $ab\mid 2^{ab}+1$.
Proof. Obviously, $a$ and $b$ are odd.
$$2^{ab}\equiv (-1)^b\equiv -1 \pmod{a}$$$$2^{ab}\equiv (-1)^a\equiv -1 \pmod{b}$$By LTE, $v_3(2^{ab}+1)=v_3(ab)+1=\alpha + \beta + 1$. Finally, as $a', b', 3^{\alpha+\beta}\mid 2^{ab}+1$ and all three numbers are pairwise coprime, $ab=a'b'3^{\alpha+\beta}\mid 2^{ab}+1$. $\square$

We construct a sequence of primes $p_2, p_3, \dots, p_{2000}$ in the following manner. Let $p_2=19$ (so that $p_2\mid 2^{3^2}+1$). Inductively assume we have constructed $p_2, p_3, \dots, p_{i}$, for some $i\ge 2$, such that:
$\bullet$ all $p_j$ with $2\le j\le i$ are distinct,
$\bullet$ $p_j>3$ and $p_j\mid2^{3^{i}}+1$ for each $2\le j\le i$.
As $2^{3^{i+1}}+1=(2^{3^i}+1)(4^{3^i}-2^{3^i}+1)$ and the greatest common divisor of the terms in the parenthesis is $3$, we conclude that $4^{3^i}-2^{3^i}+1$ has all its prime divisors distinct from $p_j$ for $2\le j\le i$. As $9\mid 2^{3^i}+1$ and $4^{3^i}-2^{3^i}+1>3$, it has a prime factor $p_{i+1}>3$, so the inductive claim holds for $i+1$ as well.

Having constructed $p_2, p_3, \dots, p_{2000}$, let $n_i=3^i p_i$ for each $2\le i\le 2000$. By LTE, $v_3(2^{n_i}+1)=i+1$. By Fermat's Little Theorem, $2^{n_i}\equiv 2^{3^i p_i}\equiv 2^{3^i}\equiv -1\pmod{p_i}$. Therefore, $n_i = 3^i p_i\mid 2^{n_i} + 1$ for each $i$. Let $n=n_2n_3\dots n_{2000}$. By repeated application of the lemma, $n\mid 2^n + 1$, and by construction, $n$ has $2000$ distinct prime divisors.
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smileapple
1010 posts
#65
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Write $x\sim y$ if $p\mid x$ iff $p\mid y$.

Let $n$ be some odd integer such that $n>3$, $n\mid 2^n+1$, and $n\nsim 2^n+1$. We claim that there exists a prime $p\nmid n$ for which $np\mid 2^{np+1}$ and $np\nsim 2^{np}+1$. Indeed, take some $p\nmid n$ and $p\mid 2^n+1$. Note that $2^{np}+1\equiv 2^n+1\equiv0\pmod p$, and since $p$ is odd, we also have $n\mid 2^{np}+1$. Hence $np\mid 2^{np}+1$. Furthermore, since $n>3$, there exists some prime $q$ satisfying $q\nmid 2^n+1$ and $q\mid 2^{np}+1$ by Zsigmondy.

Applying the above claim $1999$ times on $n=9$ implies that the answer is $\fbox{\text{yes}}$. $\blacksquare$
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amapstob
19 posts
#66
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The answer is yes.
Lemma. For all primes $p>3$, $2^p+1$ has a prime divisor greater than $p$.
Proof. $2^p\equiv -1\pmod{q}\implies 2^{\gcd(q-1,2p)}\equiv 1\pmod{q}$. If $\gcd(q-1,2p)=2$, then $q=3$, so we have to rule out $2^p+1$ being a power of three, which only happens with $p=3$ by Mihailescu's theorem. Since $p>3$, there exists $q\neq 3$ dividing $2^p+1$. Then $p\mid q-1\implies p<q$, as desired. $\blacksquare$

Claim. If $n\mid 2^{n}+1$ and $n$ is odd and has a prime divisor greater than $3$, there exists a prime with $p\mid 2^n+1$ and $p\nmid n$ such that $np\mid 2^{np}+1$.
Proof. Let $q$ be the greatest prime divisor of $n$. Then $2^q+1\mid 2^n+1$. But $2^q+1$ has a prime divisor $p$ greater than $q$ by the above lemma. So $p\nmid n$ and $p\mid 2^n+1$. Then since $p\mid 2^n+1$ and $n\mid 2^n+1$ and $n,p$ are coprime, $np\mid 2^n+1$. But $2^n+1\mid 2^{np}+1$, so we're done. $\blacksquare$

Now observe that $3^2\cdot 19 \mid 2^{3^2\cdot 19}+1$, so applying the lemma above $1998$ times finishes. $\blacksquare$
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cursed_tangent1434
618 posts
#67
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We claim that the answer is yes. In fact we can show the more general statement that for any positive integer $d$ there exists some positive integer $n$ for which $n$ has exactly $d$ distinct prime divisors and $n \mid 2^n+1$. To do this, we employ induction.

First note that $3 \mid 2^3+1$. Now, say there exists a positive integer $n_r = 3^{r}\cdot p_1p_2 \dots p_r$ for which $n_r \mid 2^{n_r}+1$. Consider
\[2^{3n_r}+1=2^{3^{r+1}\cdot p_1p_2\dots p_{r}} +1\]Now, by Zsigmondy's Theorem there exists a prime $p_{r+1} \mid 2^{3^{r+1}\cdot p_1p_2\dots p_{r}} +1$ but $p_{r+1} \nmid 2^{3^r \dot p_1p_2\dots p_r}$. Thus, this prime factor $p_{r+1} \not \in \{p_1,p_2,\dots , p_r\}$. Further,
\[3n_r \mid 2^{n_r}+1 \mid 2^{3n_r\cdot p_{r+1}}+1\]since by Lifting the Exponent Lemma, $\nu_3(2^{n_r}+1)= \nu_3(3)+\nu_3(n_r) = \nu_3(3n_r)$. Finally,
\[2^{3n_r\cdot p_{r+1}}+1 \equiv 2^{3n_r}+1 \equiv 0 \pmod{p_{r+1}}\]by construction. Thus, $3n_r\cdot p_{r+1} \mid 2^{3n_r\cdot p_{r+1}}+1$ so we can let $n_{r+1}=3n_r\cdot p_{r+1}$ which completes the induction and proves the result.
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ray66
35 posts
#68
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We will prove the result by induction.

First take the base case $n_1=9$ so that $9$ divides $513$. Now consider the number $n_2=n_1p_2$ where $p_2$ is a unique prime number dividing $2^{n_1}+1$. We know that such a $p$ exists by Zsigmondy. Therefore $2^{n_2}+1$ is also divisible by $p_2$, so we finish the induction.
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Ilikeminecraft
614 posts
#69
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I claim that there \boxed{\text{exists}} a number that has $n$ distinct prime numbers that satisfies our conditions.

Let $k_n = 9 \cdot \prod\limits_{i = 1}^{n - 1} p_i$ be a construction for $n$. I will prove this with induction.

Clearly, $k_1 = 9$ is a construction for $n = 1.$

Now, assume that $n = l$ has a valid construction. Let $p_{l}$ be a prime dividing $2^{k_l} + 1$ such that $(p_l, k_l).$ I will prove the existence of such a $p_l:$

Notice that $k$ is not even. We have that $\nu_3(2^{k_l} + 1) = \nu_3(k_l) + \nu_3(3) = 1 + 2 = 3,$ and $$\nu_{p_i}(2^{k_l} + 1) = \nu_{p_i}\left(\left(2^{k_{i + 1}}\right)^{\frac{k_l}{k_{i + 1}}} + 1\right) = \nu_{p_i}(2^{k_{i + 1}} + 1) + \nu_{p_i}\left(\prod_{j = i + 1}^{l - 1}p_j\right) = 1$$Thus, since $3k_{l} < 2^{k_l} + 1,$ there must exist a non-3 value greater than 1 that divides $\frac{2^{k_l} + 1}{k_l}.$ By picking a $p_i$ that divides that, we can gaurantee $p_i$ exists and is relatively prime to all other primes in $k_l.$

Finally, I claim that $k_{l + 1} = k_lp_{l}$ is a valid construction for $n = l + 1.$ We have that $k_l \mid 2^{k_l} + 1 \mid 2^{k_l p_l} + 1,$ and $p_l \mid 2^{k_lp_l} + 1\implies k_{l + 1} = k_lp_l \mid 2^{k_l p_l} + 1 = 2^{k_{l + 1}} + 1.$

Thus, we are done.
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