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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year 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
Nov 1, 2024
0 replies
Maximization
Butterfly   1
N 2 minutes ago by lbh_qys

Assume $a,b>0$ satisfy $a^2+b^2+\frac{1}{a}+\frac{1}{b}=\frac{27}{4}.$ Prove $2a+\frac{a}{4b}\le 5.$
1 reply
+2 w
Butterfly
32 minutes ago
lbh_qys
2 minutes ago
Strange Conditional Sequence
MarkBcc168   20
N 3 minutes ago by cushie27
Source: APMO 2019 P2
Let $m$ be a fixed positive integer. The infinite sequence $\{a_n\}_{n\geq 1}$ is defined in the following way: $a_1$ is a positive integer, and for every integer $n\geq 1$ we have
$$a_{n+1} = \begin{cases}a_n^2+2^m & \text{if } a_n< 2^m \\ a_n/2 &\text{if } a_n\geq 2^m\end{cases}$$For each $m$, determine all possible values of $a_1$ such that every term in the sequence is an integer.
20 replies
MarkBcc168
Jun 11, 2019
cushie27
3 minutes ago
Looking for a nice proof
KhuongTrang   60
N 6 minutes ago by KhuongTrang
Source: own
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that$$\color{blue}{a^2+b^2+c^2+3\ge 2\sqrt{2(a^3b+b^3c+c^3a)+3abc}.}$$
60 replies
KhuongTrang
Jul 5, 2024
KhuongTrang
6 minutes ago
Problem 6
termas   66
N 34 minutes ago by Mathandski
Source: IMO 2016
There are $n\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.

(a) Prove that Geoff can always fulfill his wish if $n$ is odd.

(b) Prove that Geoff can never fulfill his wish if $n$ is even.
66 replies
termas
Jul 12, 2016
Mathandski
34 minutes ago
No more topics!
Prove that 4p-3 is a square - Iran NMO 2005 - Problem1
sororak   20
N Nov 16, 2024 by SimplisticFormulas
Let $n,p>1$ be positive integers and $p$ be prime. We know that $n|p-1$ and $p|n^3-1$. Prove that $4p-3$ is a perfect square.
20 replies
sororak
Sep 21, 2010
SimplisticFormulas
Nov 16, 2024
Prove that 4p-3 is a square - Iran NMO 2005 - Problem1
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sororak
337 posts
#1 • 4 Y
Y by mathematicsy, RedFlame2112, Adventure10, Mango247
Let $n,p>1$ be positive integers and $p$ be prime. We know that $n|p-1$ and $p|n^3-1$. Prove that $4p-3$ is a perfect square.
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Goutham
3130 posts
#2 • 3 Y
Y by RedFlame2112, Adventure10, Mango247
Here
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Rust
5049 posts
#3 • 3 Y
Y by RedFlame2112, Adventure10, Mango247
$n=1$ solution for any $p$. Therefore I think condition is:
Let $n>1$ (not $p>1$) integer and $p$ prime.

Because $1<n<p$ $p|n^3-1=(n-1)(n^2+n+1)\to p|n^2+n+1$ and $n|p-1\to p=kn+1,k\le n+1$.
If $k<n+1$, then $p|n^2+n+1-p=n(n+1-k)$ contradition with $p>n$ and prime. Therefore $p=n^2+n+1$ and $4p-3=(2n+1)^2.$
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Binomial-theorem
3959 posts
#4 • 4 Y
Y by AlastorMoody, mst.4921, RedFlame2112, Adventure10
Once you arrive at $p=kn+1$ you can also observe that:

\begin{eqnarray*}  p\mid n^2+n+1\mid kn^2+kn+k \\ kn+1\mid \left(kn^2+kn+k\right)-n\left(kn+1\right) \\ kn+1\mid kn+k-n \\ k-n\ge 1\implies k=n+1\end{eqnarray*}

The finish is the same as in Rust's.
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kootrapali
4527 posts
#5 • 2 Y
Y by RedFlame2112, Adventure10
From the first divisibility, we know $p-1\geq n\rightarrow p\geq n+1$. Since $p|n^3-1\rightarrow p|(n-1)(n^2+n+1)$, and $p\geq n+1>n-1$, since $p$ is prime we have $\gcd (p,n-1)=1$, so $p|n^2+n+1$. Let $k$ be a positive integer such that $pk=n^2+n+1$. Taking this modulo $n$, we have $pk\equiv k\equiv1$ mod $n$. Now let $k=an+1$ for non-negative $a$. Then $p(an+1)=n^2+n+1\rightarrow apn+p=n^2+n+1$. Since $p\geq n+1$ and for $a\geq 1$, we have $apn>n^2$, we must have $a=0$, or $k=1$. Then $p=n^2+n+1$, which also satisfies the first divisibility. Finally, we have $4p-3=4n^2+4n+1=(2n+1)^2$, completing the proof.
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Albert123
204 posts
#6 • 1 Y
Y by RedFlame2112
Here is my solution:
Since: $n|p-1 \implies p=nk+1$ for some $k$ positive integer
Of the condition: $p|(n^3-1) \implies p|(n-1)(n^2+n+1) \implies p|n-1$ or $p|n^2+n+1$
if $p|n-1$ $\implies p \le n-1$ but of the condition $n|p-1 \implies n \le p-1$
i.e $p \le n-1$ and $n \le p-1$. Then $p+1 \le n \le p-1$ so it's a contradiction
Thus: $p|n^2+n+1 \implies p \le n^2+n+1$
Replacing: $nk+1|n^2+n+1$
$\implies nk+1|(n^2+n+1)k=n^2k+nk+k$
$\implies nk+1|n^2k+nk+k-n(nk+1)=nk+k-n$
$\implies nk+1 \le nk+k-n \implies n+1 \le k$
$\implies n(n+1)+1\le nk+1=p$
$n^2+n+1\le p$ but $p \le n^2+n+1$
$\implies p=n^2+n+1$
Thus: $4p-3=4n^2+4n+1=(2n+1)^2$.$\blacksquare$
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Taco12
1757 posts
#7 • 1 Y
Y by RedFlame2112
We can manipulate the conditions as follows:
$n|p-1 \implies p=nk+1$ and $p\geq n+1$.
$p|n^3-1 \leftrightarrow p|(n-1)(n^2+n+1) \implies$ either $p|n-1$ or $p|n^2+n+1$.
But $p|n-1$ is not possible because $p\geq n+1$, so $p|n^2+n+1$. Hence, $p \leq n^2+n+1$.

Additionally, $nk+1|kn^2+kn+k$, so $nk+1|nk+k-n$. Thus, $k \geq n+1$. Substituting $p=nk+1$ back in, we find $p\geq n^2+n+1$. However, we also know that $p \leq n^2+n+1$. Therefore, $p=n^2+n+1$, so $4p-3=4n^2+4n+1=(2n+1)^2$, which completes the problem.
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megarnie
5325 posts
#8 • 1 Y
Y by RedFlame2112
Let $p=nk+1$.

We have $p\mid n^3-1=(n-1)(n^2+n+1)$. Now $p>n$, so $p\nmid n-1$, which implies $p\mid n^2+n+1$.

So $nk+1\mid n^2+n+1$, which implies $nk+1\mid kn^2+kn+k-n(nk+1)=kn+k-n$.

Thus, $nk+1\le nk+k-n\implies n+1\le k$. Thus, $n^2+n+1=n(n+1)+1\le nk+1=p$. So $p=n^2+n+1$.

This implies $4p-3=4n^2+4n+1=(2n+1)^2$, as desired.
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Arr0w
2908 posts
#9 • 1 Y
Y by RedFlame2112
Notice that
$$n~|~p-1\implies n\le p-1\implies n+1\le p\implies p=nk+1$$for some $k$ and that
$$p~|~(n-1)(n^2+n+1)\implies p\le n^2+n+1$$since $p~|~n-1$ is impossible in this case. Playing around with divisibility some more gives
$$p~|~n^2+n+1~|~kn^2+kn+k\implies nk+1~|~kn^2+kn+k\implies nk+1~|~kn^2+kn+k-n(nk+1)$$which gives
$$nk+1~|~kn+k-n\implies nk+1\le kn+k-n\implies n+1\le k.$$Using what we gathered previously, mainly that $p=nk+1$, we see that
$$n(n+1)+1\le nk+1\implies n^2+n+1\le p$$which means we have
$$n^2+n+1\le p\le n^2+n+1\implies p=n^2+n+1.$$So using this we see that
$$4p-3=4(n^2+n+1)-3=(2n+1)^2$$which proves the desired $\square$
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Mogmog8
1080 posts
#10 • 1 Y
Y by centslordm
Let $p=\ell n+1$ and notice $p\mid (n-1)(n^2+n+1)$ so $p\mid n-1$ or $p\mid n^2+n+1.$ We see the former implies $n+p\le n+p-2,$ a contradiction; hence, $p\mid n^2+n+1.$ Then, $$\ell n+1\mid \ell(n^2+n+1)-n(\ell n+1)=(\ell-1)n+\ell>0.$$Thus, \begin{align*}\ell n+1\le (\ell-1)n+\ell&\implies n+1\le \ell\\&\implies p\ge (n+1)n+1=n^2+n+1\\&\implies p=n^2+n+1\\&\implies 4p-3=4n^2+4n+1=(2n+1)^2.\end{align*}$\square$
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mathlogician
1051 posts
#11 • 2 Y
Y by GeoKing, Mango247
Write $p=kn+1$, and note that $p \mid (n-1)(n^2+n+1)$. Since $p > n-1$, it follows that $p \mid n^2+n+1$. Thus, for some integer $t$ we have that $t(kn+1) = n^2+n+1$, and noting that as both $kn+1$ and $n^2+n+1$ are congruent to $1 \pmod n$, $t$ must be congruent to $1 \pmod n$ as well. However, if $t \geq n+1$ then the left hand side of $t(kn+1) \geq (n+1)^2 > n^2+n+1$ which contradicts the divisibility condition, so $t=1$ and $p = n^2+n+1$. Finally, $4p-3 = 4(n^2+n+1)-3 = 4n^2+4n+1 = (2n+1)^2$, as desired.
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HamstPan38825
8684 posts
#12
Y by
You have to love the hilarious answer extraction :)

For size reasons we must have $p \mid n^2+n+1$. But then $np \mid n^2+n-p+1$; on the other hand $$np \geq n(n+1) > n^2+n-p+1$$so it follows $p = n^2+n+1$ and $4p-3 = (2n+1)^2$ is a perfect square.
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ArmandoYunke
5 posts
#13
Y by
Is this problem have solution with LTE??
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lifeismathematics
1188 posts
#14
Y by
Cute! , writing for storage purpose.
$p|n^3-1 \implies p|(n-1)(n^2+n+1)$ , now if $p|n-1\implies p|n-1 , n|p-1 \implies p\geqslant n-1 \hspace{0.2cm}  \text{and} \hspace{0.2cm}  p \leqslant n-1$ that's a contradiction for a prime $p$. Hence $p|n^2+n+1 \implies p \leqslant n^2+n+1 \hspace{0.3cm} (\star)$

Now , since $n|p-1 \implies p=nt+1$ , for some $t \in \mathbb{Z}^{+}$

so we must have $$nt+1|n^2+n+1 \implies nt+1| n^2+n+1-(nt+1)^2 \implies nt+1| n(n-nt^2+1-2t)$$, since $\gcd(nt+1,n)=1$ we must have $nt+1|n-nt^2+1-2t \implies nt+1|n+1-t$ , now we also notice that if $n+1-t>0$ , then $$nt+1<n+1-t \implies n(t-1) \leqslant -t$$which is not posisble , hence $n+1 \leqslant t$ , which gives $n^2+n+1 \leqslant p \leqslant n^2+n+1 \implies p=n^2+n+1 \implies 4p-3=(2n+1)^2$. $\square$
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de-Kirschbaum
164 posts
#15
Y by
Note that $n|p-1 \implies p=nk+1$ and $p|n^3+1 \implies p|(n+1)(n^2+n+1) \implies p|n^2+n+1$. Note now that $p|k(n^2+n+1)-k(nk+1)=nk(n+1-k)$. Since $\gcd(nk+1,nk)=1$, we have that $nk+1|n+1-k$, but $n+1-k \leq nk+1$ so it must be negative or 0, thus $n+1-k \leq 0 \implies k \geq n+1$. However, if $k=n+1$ then $p=n(n+1)+1=n^2+n+1$, so that is actually the only value $k$ can take. That means $p=n^2+n+1, 4p-3=4n^2+4n+1=(2n+1)^2$ and we see that it is a perfect square.
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KevinYang2.71
349 posts
#16
Y by
Since $p\mid (n-1)(n^2+n+1)$, we must have $p\mid n^2+n+1$ because $p>n-1$. Let $p=:an+1$ and $n^2+n+1=:pd$ for positive integers $a$ and $d$. Then $n^2+n+1=d(an+1)$. Taking mod $n$ gives $d\equiv 1\pmod{n}$ so $d=:bn+1$ for $b\in\mathbb{N}$. It follows that $abn+a+b=n+1$ so $n=\frac{1-a-b}{ab-1}$. If $ab-1$ is positive then $1-a-b$ is negative, which makes $n$ negative. Thus $ab-1$ is negative so $b=0$. Hence $d=1$ so $4d-3=(2n+1)^2$, as desired. $\square$
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Flint_Steel
36 posts
#17
Y by
From $n|p-1$, we get $n+1\leq p $. Then from the fact that $p|(n-1)(n^{2}+n+1)$, we get $p|n^{2}+n+1 \Rightarrow p \leq n^{2}+n+1$.
Let $k$ be an positive integer such that $nk+1=p$. So, $nk+1 \leq n^{2}+n+1 \Rightarrow k \leq n+1$. Then $p=nk+1 \geq n^{2}+n+1 $.
Which means, $p=n^{2}+n+1$, then it is easy to see that $4p-3$ is a perfect square.
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Ywgh1
118 posts
#18
Y by
Iran 2005

Obviously we have that $p|n^2+n+1$, so we have that.
$np \mid n^2+n-p+1$, but also.
$$np \geq n(n+1) > n^2+n-p+1$$
Hence $p=n^2+n+1$, hence we are done.
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Math_456
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This post has been edited 1 time. Last edited by Math_456, Aug 18, 2024, 10:13 AM
Reason: oops
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John_Mgr
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$n|p-1$. So, $p=nk+1$
Also,$p-1 \geq n $ and $p \geq n+1$
$p|(n-1)(n^{2}+n+1)$. If $p|n-1$ then $n-1 \geq p$.
However, $n-1 \geq p \geq n+1$ not possible, So $p|n^{2}+n+1$ and $p \leq n^{2}+n+1 \hspace{0.3cm} (\star)$.
$p|n^{2}+n+1$, $nk+1|n^{2}+n+1$, $nk+1|k(n^{2}+n+1)-n(kn+1)$ $\Rightarrow$ $nk+1|nk+k-n$ So,$ nk+k-n \geq kn+1$, $n+1 \leq k$.
we have $p =nk+1 \geq n(n+1)+1$ then $p \geq n^{2}+n+1 \hspace{0.3cm} (\star)$
From the $\hspace{0.cm} (\star)$ We got, $p=n^{2}+n+1$. , $4p-3=(2n+1)^{2}$
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SimplisticFormulas
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Cool!

Let $n^2 +n+1=kp$, where $k \in \mathbb{N}$
$\implies n(n+1)=kp-1$
$\implies n \mid kp-1$
Combining with $n \mid p-1$, we get $ n \leq gcd(kp-1,p-1)$
Note that if $2 \leq k$, then $gcd(kp-1,p-1)=1$, which would imply $n=1$ amd $p=3$.
Hence, for rest of the primes, $k=1$
$\implies n^2 +n+1=p$
$\implies n^2+n+(1-p)=0$
Since this quadratic in $n$ has integer roots, the discriminant, that is $1-4(1-p)=4p-3$ mus be a perfect square. $\blacksquare$
[comment] This took way longer than it should have duh [/comment]
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