Prove that
such that
Y by
If 
prove that 
prove that 
This post has been edited 1 time. Last edited by Marinchoo, Nov 17, 2020, 2:54 PM
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Number Theory
MuradSafarli 4
N
by mdnajibl477
find all natural numbers 
such that the following equation holds:
![\[
7^a + 1 = 2b^2
\]](//latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
4 replies
euler-totient function
Laan 0
Proof that there are infinitely many positive integers 
such that
such that
0 replies
Cubic function from Olymon
Adywastaken 3
N
by Ahmed_mallek
Source: Olymon Volume 11 2010 663
Find all 
such that
such that
3 replies
+1 wColouring numbers
kitun 2
N
by quasar_lord
What is the least number required to colour the integers 
such that for any set of consecutive integers taken from the given set of integers, there will always be a colour colouring exactly one of them? That is, for all integers 
such that 
, there will be a colour coloring exactly one integer from the set 
.
such that for any set of consecutive integers taken from the given set of integers, there will always be a colour colouring exactly one of them? That is, for all integers 
such that 
, there will be a colour coloring exactly one integer from the set 
.
2 replies
Mathhhhh
mathbetter 4
N
by Amkan2022
Three turtles are crawling along a straight road heading in the same
direction. "Two other turtles are behind me," says the first turtle. "One turtle is
behind me and one other is ahead," says the second. "Two turtles are ahead of me
and one other is behind," says the third turtle. How can this be possible?
direction. "Two other turtles are behind me," says the first turtle. "One turtle is
behind me and one other is ahead," says the second. "Two turtles are ahead of me
and one other is behind," says the third turtle. How can this be possible?
4 replies
Finally hard NT on UKR MO from NT master
mshtand1 3
N
by Jupiterballs
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 11.4
A pair of positive integer numbers 
is given. It turns out that for every positive integer number 
, for which the numbers 
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
Proposed by Oleksii Masalitin
is given. It turns out that for every positive integer number 
, for which the numbers 
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?Proposed by Oleksii Masalitin
3 replies
Sequence bounding itself
juckter 2
N
by Math-Problem-Solving
Source: Own
We say a sequence of integers 
is self-bounded if for each 
, 
there exist at least 
terms of the sequence that are less than or equal to . Find the maximum possible value of 
for a self-bounded sequence .
is self-bounded if for each 
, 
there exist at least 
terms of the sequence that are less than or equal to . Find the maximum possible value of 
for a self-bounded sequence .
2 replies
Checkerboard
Ecrin_eren 0
On an 8×8 checkerboard, what is the minimum number of squares that must be marked (including the marked ones) so that every square has exactly one marked neighbor? (We define neighbors as squares that share a common edge, and a square is not considered a neighbor of itself.)
0 replies
Inequality
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Y by
If a, b and c can be negative, the inequality is wrong ( a=1 b = -1/2. c = -2)
If a, b, c are positive reals, we have by hypothesis:
cc = 1/a. ac = 1/b ab = 1/c. So
(a^2 + bc)(b^2 + ac)(c^2 + ab) =
(a^2 + 1/a)(b^2 + 1/b)(c^2 + 1/b) =
(a^3 + 1)(b^3 + 1)(c^3 + 1) because abc=1
And
(a^3 + 1)/2. (b^3 + 1)/2. (c^3 + 1) /2
=< (a^3 . b^3. c^3 + 1)/2 =1
by using Tchebychev
Also, (a^3 + b^3 + c^3)/3 >= abc=1>=
(a^3 + 1)/2. (b^3 + 1)/2. (c^3 + 1) /2
That finishes the proof. Egality hold if a=b=c= 1.
If a, b, c are positive reals, we have by hypothesis:
cc = 1/a. ac = 1/b ab = 1/c. So
(a^2 + bc)(b^2 + ac)(c^2 + ab) =
(a^2 + 1/a)(b^2 + 1/b)(c^2 + 1/b) =
(a^3 + 1)(b^3 + 1)(c^3 + 1) because abc=1
And
(a^3 + 1)/2. (b^3 + 1)/2. (c^3 + 1) /2
=< (a^3 . b^3. c^3 + 1)/2 =1
by using Tchebychev
Also, (a^3 + b^3 + c^3)/3 >= abc=1>=
(a^3 + 1)/2. (b^3 + 1)/2. (c^3 + 1) /2
That finishes the proof. Egality hold if a=b=c= 1.
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#5
Apr 29, 2020, 12:24 AM
Y by
Taking 
shows that it cannot hold.
In general, taking
shows that it is false for sufficiently large 
while the reverse is false for sufficiently small .
shows that it cannot hold.In general, taking
shows that it is false for sufficiently large 
while the reverse is false for sufficiently small .This post has been edited 1 time. Last edited by Victoria_Discalceata1, Apr 29, 2020, 12:35 AM
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true for all positive reals 
,
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