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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inspired by hunghd8
sqing   0
8 minutes ago
Source: Own
Let $ a,b,c\geq 0 $ and $ a+b+c\geq 2+abc . $ Prove that
$$a^2+b^2+c^2-\frac{1}{2}a^2b^2c^2\geq 2$$
0 replies
sqing
8 minutes ago
0 replies
Find min
hunghd8   3
N 23 minutes ago by sqing
Let $a,b,c$ be nonnegative real numbers such that $ a+b+c\geq 2+abc $. Find min
$$P=a^2+b^2+c^2.$$
3 replies
hunghd8
4 hours ago
sqing
23 minutes ago
Interesting inequality
sqing   5
N an hour ago by sqing
Source: Own
Let $ a,b >0. $ Prove that
$$  \frac{1}{\frac{a}{a+b}+\frac{a}{2b}} +\frac{1}{\frac{b}{a+b}+\frac{1}{2}} +\frac{a}{2b} \geq \frac{5}{2}  $$
5 replies
sqing
Feb 26, 2025
sqing
an hour ago
sum of divisors nt
Soupboy0   0
an hour ago
Source: own
Let $\epsilon(n)$ denote the sum of the sum of the factors of all positive $\mathbb Z \le n$, for example, $\epsilon(5) $ is the sum of the factors of $5$ added to the sum of the factors of $4$ and so on until the sum of the factors of $1$, which would be $(1+5)+(1+2+4)+(1+3)+(1+2)+(1) = 21$. Let $M(n)$ denote $\sum_{i=1}^{n} n \pmod{i}$. Show that $\epsilon(n) + M(n) = n^2$ or find a counterexample
0 replies
Soupboy0
an hour ago
0 replies
No more topics!
All-Russian Olympiads 2002 p2
mousavi   4
N Mar 24, 2011 by jgnr
A monic quadratic polynomial $f$ with integer coefficients attains prime values at three consecutive integer points.show that it attains a prime value at some other integer point as well.
4 replies
mousavi
Mar 22, 2011
jgnr
Mar 24, 2011
All-Russian Olympiads 2002 p2
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mousavi
222 posts
#1 • 1 Y
Y by Adventure10
A monic quadratic polynomial $f$ with integer coefficients attains prime values at three consecutive integer points.show that it attains a prime value at some other integer point as well.
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tenniskidperson3
2376 posts
#2 • 2 Y
Y by Adventure10, Mango247
We know that if $f(x)=x^2-bx+c$ then $f(a)=f(b-a)$. Thus if the consecutive integers $x-1, x, x+1$ are not $\frac{b}{2}-1, \frac{b}{2}, \frac{b}{2}+1$ then one of the three consecutive numbers is at least a distance of $\frac{3}{2}$ from $\frac{b}{2}$. Let this number be $y$. Then

$f(y)=f(b-y)$ and $|y-(b-y)|=|2y-b|=\left|2\left(y-\frac{b}{2}\right)\right|\geq2\left(\frac{3}{2}\right)=3$.

Thus this number cannot be included in the 3 consecutive integers because it is 3 away from one of them. So we have another integer input to the function with a prime output.

So now we assume that the 3 numbers are in fact $\frac{b}{2}-1, \frac{b}{2}, \frac{b}{2}+1$. These are integers only if $b=2k$ or $f(x)=x^2-2kx+c=(x-k)^2+(c-k^2)$. We know that at $k-1, k, k+1$ these are primes. But

$f(k+1)-f(k)=((k+1)-k)^2+(c-k^2)-(k-k)^2-(c-k^2)=1$.

The only two primes that are 1 apart are $2$ and $3$. Thus $f(k)=2$, so $c-k^2=2$ and $f(x)=(x-k)^2+2$. Then $f(k+3)=(k+3-k)^2+2=11$ which is another prime value at an integer point.
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modularmarc101
2208 posts
#3 • 1 Y
Y by Adventure10
tenniskidperson3 wrote:
... Let this number be $y$. Then

$f(y)=f(b-y)$ and $|y-(b-y)|=|2y-b|=\left|2\left(y-\frac{b}{2}\right)\right|\geq2\left(\frac{3}{2}\right)=3$.

Thus this number cannot be included in the 3 consecutive integers because it is 3 away from one of them. ...

Why must $b-y$ be one of the 3 consecutive integers?
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tenniskidperson3
2376 posts
#4 • 2 Y
Y by Adventure10, Mango247
There is no reason why it should be; in fact it isn't. I don't understand what you're saying; $b-y$ is at least 3 away from $y$ which is one of the consecutive numbers, so $b-y$ cannot be.
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jgnr
1343 posts
#5 • 2 Y
Y by Adventure10, Mango247
Suppose $f(n-1),f(n),f(n+1)$ are primes. Define another polynomial $g(x)=f(x+n)$. So $g$ is a monic quadratic polynomial with integer coefficients such that $g(-1),g(0),g(1)$ are primes. It is enough to show that $g(x)$ attains a prime value at another integer point. Let $g(x)=x^2+ax+b$. Note that $g(x)=g(-a-x)$. So $g(-a-1),g(-a),g(-a+1)$ are primes. If the set $\{-a-1,-a,-a+1\}$ does not coincide with $\{-1,0,1\}$, then we're done. Otherwise, we get $a=0$. Then $g(-1)=1+b$ and $g(0)=b$, both of which are primes. So $b=2$ and $g(x)=x^2+2$, which attains a prime value at $g(3)=11$.
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