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Number theory for people who love theory
Assassino9931   2
N 5 minutes ago by MathLuis
Source: Bulgaria RMM TST 2019
Prove that there is no positive integer $n$ such that $2^n + 1$ divides $5^n-1$.
2 replies
Assassino9931
Jul 31, 2024
MathLuis
5 minutes ago
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sum of 4 primes with 5 <p <q <r <s <p + 10 is divisible by 60
parmenides51   4
N an hour ago by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 4
Let $p, q, r$ and $s$ be four prime numbers such that $$5 <p <q <r <s <p + 10.$$Prove that the sum of the four prime numbers is divisible by $60$.

(Walther Janous)
4 replies
parmenides51
Dec 18, 2020
MathIQ.
an hour ago
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2player game, adding numbers, whoever reaches no >= 2019 wins
parmenides51   2
N 2 hours ago by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 3
Alice and Bob are playing a year number game.
There will be two game numbers $19$ and $20$ and one starting number from the set $\{9, 10\}$ used. Alice chooses independently her game number and Bob chooses the starting number. The other number is given to Bob. Then Alice adds her game number to the starting number, Bob adds his game number to the result, Alice adds her number of games to the result, etc. The game continues until the number $2019$ is reached or exceeded.
Whoever reaches the number $2019$ wins. If $2019$ is exceeded, the game ends in a draw.
$\bullet$ Show that Bob cannot win.
$\bullet$ What starting number does Bob have to choose to prevent Alice from winning?

(Richard Henner)
2 replies
parmenides51
Dec 18, 2020
MathIQ.
2 hours ago
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60^o angle wanted, equilateral on a square
parmenides51   4
N 2 hours ago by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 2
A square $ABCD$ is given. Over the side $BC$ draw an equilateral triangle $BCS$ on the outside. The midpoint of the segment $AS$ is $N$ and the midpoint of the side $CD$ is $H$. Prove that $\angle NHC = 60^o$.
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(Karl Czakler)
4 replies
parmenides51
Dec 18, 2020
MathIQ.
2 hours ago
J
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(x^2 + y^2)/(x + y)= 10 diophantine
parmenides51   7
N 2 hours ago by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 1
Let $x$ and $y$ be integers with $x + y \ne 0$. Find all pairs $(x, y)$ such that $$\frac{x^2 + y^2}{x + y}= 10.$$
(Walther Janous)
7 replies
parmenides51
Dec 18, 2020
MathIQ.
2 hours ago
J
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Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   9
N 2 hours ago by hectorleo123
Source: Serbian selection contest for the IMO 2025
Let $ABC$ be an acute triangle. Let $A'$ be the reflection of point $A$ over the line $BC$. Let $O$ and $H$ be the circumcenter and the orthocenter of triangle $ABC$, respectively, and let $E$ be the midpoint of segment $OH$. Let $D$ and $L$ be the points where the reflection of line $AA'$ with respect to line $OA'$ intersects the circumcircle of triangle $ABC$, where point $D$ lies on the arc $BC$ not containing $A$. If \( M \) is a point on the line \( BC \) such that \( OM \perp AD \), prove that \( \angle MAD = \angle EAL \).

Proposed by Strahinja Gvozdić
9 replies
OgnjenTesic
Yesterday at 4:02 PM
hectorleo123
2 hours ago
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collinear wanted, regular hexagon
parmenides51   3
N 2 hours ago by MathIQ.
Source: 2023 Austrian Mathematical Olympiad , Junior Regional Competition , Problem 2
Let $ABCDEF$ be a regular hexagon with sidelength s. The points $P$ and $Q$ are on the diagonals $BD$ and $DF$, respectively, such that $BP = DQ = s$. Prove that the three points $C$, $P$ and $Q$ are on a line.

(Walther Janous)
3 replies
parmenides51
Mar 26, 2024
MathIQ.
2 hours ago
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(x + y)(y + z)(z + x)/{xyz} if (x+ y}/z=(y + z)/x=(z + x)/y
parmenides51   8
N 2 hours ago by MathIQ.
Source: 2023 Austrian Mathematical Olympiad, Junior Regional Competition , Problem 1
Let $x, y, z$ be nonzero real numbers with $$\frac{x + y}{z}=\frac{y + z}{x}=\frac{z + x}{y}.$$Determine all possible values of $$\frac{(x + y)(y + z)(z + x)}{xyz}.$$
(Walther Janous)
8 replies
parmenides51
Mar 26, 2024
MathIQ.
2 hours ago
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a! + b! = 2^{c!}
parmenides51   7
N 2 hours ago by MathIQ.
Source: 2023 Austrian Mathematical Olympiad, Junior Regional Competition , Problem 4
Determine all triples $(a, b, c)$ of positive integers such that
$$a! + b! = 2^{c!}.$$
(Walther Janous)
7 replies
parmenides51
Mar 26, 2024
MathIQ.
2 hours ago
J
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Geometric inequality with angles
Amir Hossein   7
N 2 hours ago by MathIQ.
Let $p, q$, and $r$ be the angles of a triangle, and let $a = \sin2p, b = \sin2q$, and $c = \sin2r$. If $s = \frac{(a + b + c)}2$, show that
\[s(s - a)(s - b)(s -c) \geq 0.\]
When does equality hold?
7 replies
Amir Hossein
Sep 1, 2010
MathIQ.
2 hours ago
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IMO 2014 Problem 3
v_Enhance   103
N 2 hours ago by Mysteriouxxx
Source: 0
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[ \angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
103 replies
v_Enhance
Jul 8, 2014
Mysteriouxxx
2 hours ago
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Equation with powers
a_507_bc   6
N Apr 3, 2025 by EVKV
Source: Serbia JBMO TST 2024 P1
Find all non-negative integers $x, y$ and primes $p$ such that $$3^x+p^2=7 \cdot 2^y.$$
6 replies
a_507_bc
May 25, 2024
EVKV
Apr 3, 2025
J
Equation with powers
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G H BBookmark kLocked kLocked NReply
Source: Serbia JBMO TST 2024 P1
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a_507_bc
678 posts
a_507_bc
#1 h  May 25, 2024, 5:27 PM
Y by
Find all non-negative integers $x, y$ and primes $p$ such that $$3^x+p^2=7 \cdot 2^y.$$
Z K Y
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NO_SQUARES
1133 posts
NO_SQUARES
#2 h  May 25, 2024, 6:10 PM
Y by
a_507_bc wrote:
Find all non-negative integers $x, y$ and primes $p$ such that $$3^x+p^2=7 \cdot 2^y.$$
If $y=0$ then $x=1, p=2$. If $y=1$ then by mod 3 $x=0$ and there are no solutions. Now let $y>1$, so $4 | RHS$.
Note that since $4|7-3$ we have $2 \not | x$. After this look at mod 8 to get $y<3$ ($p \not = 2$).
This post has been edited 2 times. Last edited by NO_SQUARES, May 25, 2024, 7:30 PM
Reason: was wrong
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RagvaloD
4918 posts
RagvaloD
#3 h  May 25, 2024, 7:21 PM • 2 Y
Y by NO_SQUARES, ehuseyinyigit
$p=2 \to y=0,x=1$
$p$ is odd $\to 3^x+p^2 \equiv 2,4 \pmod {8} \to y<3$
for $y=1$ there are no solutions
For $y=2: x=1,p=5$
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Assassino9931
1362 posts
Assassino9931
#4 h  May 25, 2024, 8:20 PM
Y by
We only need that $p$ is odd if $p \geq 3$. Indeed, mod 8 we have $3^x \equiv 3,1$ and $p^2\equiv 1$ and so $3^x + p^2 \equiv 2, 4$ while $7 \cdot 2^y \equiv 0$ for $y\geq 3$, contradiction. Hence either $p=2$ or $y\leq 2$.

If $p=2$, then parity insists on $y=0$, so $x=1$. If $y=2$, then only $p=5$ and $x=1$ works. If $y=1$, then there are no solutions. If $y=0$, then $p=2$ and $x=1$ works.

Hence all solutions are $(x,y,p) = (1,0,2), (1,2,5)$.
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THE_SOLVER
1 post
THE_SOLVER
#5 h  Jul 31, 2024, 4:16 PM • 1 Y
Y by JelaByteEngineer
Simply by applying mod 8 which restricts or bounds the value of y i.e y<3
By checking manually which gives us 2 solutions i.e (1,2,0);(1,2,5)
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ali123456
52 posts
ali123456
#6 h  Mar 18, 2025, 4:34 PM
Y by
sketch of my solution
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EVKV
71 posts
EVKV
#7 h  Apr 3, 2025, 2:32 AM
Y by
p is odd for $y \neq 0$
and clearly x= 1 , y= 0, p=2 satisfies
$For y \geq 3$
$either 2,4 \equiv 0$ $mod$ $8$ nonsense
now checking remaining
all solutions are $(x,y,p) = (1,0,2), (1,2,5)$.
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