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TRAN THAI HUNG
Riemann Hypothesis
Riemann Hypothesis

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#1
HCM city TST
I hope you enjoy this
1) Find the smallest integer number n>1 such that:
M = \frac{{1^2 + 2^2 + ... + n^2 }}{n} is a perfect square

2) Let sequence \{ u_n \}: u_1 = a > 1 ; u_{n + 1} = u_n^2 - u_n + 1,\forall n \ge 1
Find:
\lim (\frac{1}{{u_1 }} + \frac{1}{{u_2 }} + ... + \frac{1}{{u_n }})

3) Let \Delta ABC: AB=AC. M is a point in \Delta ABC such that \angle BMC = 90^0 + \frac{{\widehat A}}{2}. Construct the paralellogram MKBD and MHCE; K in AB, H in AC; D,E in BC. N is the intersection point of KD and HE. Find the locus of N?

4) a,b,c \in R and \left\{ \begin{array}{l} a^2 + c^2 = 10 \\ b^2 + 2b(a + c) = 60 \\ \end{array} \right.
A=b(c-a)
Find Max and Min of A

5)Find all the finited set have at least 2 elements with all the element is the positive integers.And if a,b are the two element of A(b>a), \frac{{a^2 }}{{b - a}} is also the element of A.

6)
Find all the function f: R \to R such that
f(f(x) + y) = 2x + f(f(f(y)) - x)\forall x,y \in R
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PostPosted: Sun Nov 01, 2009 1:57 am  Back to top 
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JakiroVn
P versus NP
P versus NP

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#2
 Hưng
Hưng là thằng mập!
How you can solve Problem 3 Hung ? Blush Blush
This is the second day, post the first day if you have. I hope MathVnpro would write hís solution. Smile

PostPosted: Mon Nov 02, 2009 4:21 am  Back to top 
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TRAN THAI HUNG
Riemann Hypothesis
Riemann Hypothesis

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#3
I'll post the solution but not now.Let's wait for everyone sovle it.
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PostPosted: Wed Nov 04, 2009 6:30 am  Back to top 
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ocha
Yang-Mills Theory
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#4
one


\frac {(n + 1)(2n + 1)}{6} = k^2, k\in \mathbb{N}

Let n + 1 = m and we get 2m^2 - m = 6k^2 \qquad(1)

Since m is an integer, the discriminant of (1) is a perfect square
\therefore 1 + 48k^2 = a^2 \Leftrightarrow a^2 - 48k^2 = 1

We have a pell equation, and the smallest solution is clearly a = 7, k = 1

(7 + \sqrt {48})^2 = 97 + 14\sqrt {48}, but this gives m = \frac {1 \pm a}{4} = \frac {49}{2}

(7 + \sqrt {48})^3 = 1351 + 260\sqrt {48} giving m = \frac {1 + 1351}{4} = 338

So n = 337 and \frac {(337 + 1)(2\cdot 337 + 1)}{6} = 195^2 which works

n = \boxed{337}



PostPosted: Wed Nov 04, 2009 9:48 pm  Back to top 
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Differ
Riemann Hypothesis
Riemann Hypothesis


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#5
Hints for 3
It's the circumcircle
More hints
A, M, and N are collinear

PostPosted: Fri Nov 06, 2009 12:44 am  Back to top 
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grafitti123
P versus NP
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#6
another similar solution for 1

\frac {(n + 1)(2n + 1)}{6} = k^2}
since (n + 1)(2n + 1) are coprime we get,
n + 1 = 2a^{2} and 2n + 1 = 3b^{2}
if 2a = k eliminating n we get,
k^2 - 3b^2 = 1
solving, (2,1)(7,4)(26,15)
since b is odd, we take 26,15 to get solution 337

The other case :
n + 1 = 6a^2
2n + 1 = b^2
It gives 12a^{2} = b^{2} + 1
This
has no solutions since there is no square leaving a remainder of 11 on division with 12.


PostPosted: Fri Nov 06, 2009 6:31 am  Back to top 
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darkmage2009
Hodge Conjecture
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#7
5. Partial, I might add more as I figure out more about the problem.
Click to reveal hidden content

Let A be the set of all sets S with the properties listed.
(It has at least 2 elements, all of its elements are positive integers, and if a,b\in S and b > a, then \frac {a^2}{b - a}\in S).
Part 1: All sets of the form \{k, 2k\}, k\in \mathbb{N}, are in A.
Proof: Since we are working with positive integers, 2k > k. Then the only possible choices of a and b are a = k and b = 2k. Then \frac {a^2}{b - a} = \frac {k^2}{2k - k} = k, which is already in the set, so the set has all the properties.

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1+1=2 in base 10, and 1+1=10 in base 2. Coincidence? I think not!

PostPosted: Fri Nov 06, 2009 3:53 pm  Back to top 
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ocha
Yang-Mills Theory
Yang-Mills Theory

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#8
number 2

u_{n + 1} = u_n^2 - u_n + 1

u_{n + 1} - 1 = u_n(u_n - 1)

\frac {1}{u_{n + 1} - 1} = \frac {1}{u_n(u_n - 1)} = \frac {1}{u_n - 1} - \frac {1}{u_n}

\therefore \frac {1}{u_n} = \frac {1}{u_n - 1} - \frac {1}{u_{n + 1} - 1}

So by telescoping series

\sum_{k = 1}^{\infty} \frac {1}{u_k} = \sum_{k = 1}^{\infty} \frac {1}{u_k - 1} - \frac {1}{u_{k + 1} - 1} = \frac {1}{u_1 - ...

PostPosted: Sun Nov 08, 2009 12:07 am  Back to top 
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ocha
Yang-Mills Theory
Yang-Mills Theory

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#9
number 6

f(f(x)+y) = 2x+f(f(f(y))-x) \qquad(1)

Proof that f is surjective

let y=-f(x) and rearrange to get

f(0)-2x = f[ f(f(-f(x)))-x]

Since f(0) is fixed, we can choose x such that f(0)-2x is any real number, hence f is surjective


proof that f is injective

assume f(a)=f(b) \Longrightarrow f[f(f(a))-x]+2x = f[f(f(b))-x]+2x

\therefore f(f(x)+a)=f(f(x)+b, but since f is surjective, let f(x)=y-a

\therefore f(y) = f(y+b-a) and by induction f(y)=f(y+n(b-a))

Now plugging into (1) gives
f(f(x)+y)=f(f(x+b-a)+y) = f[f(f(y))+x+b-a] + 2x + 2(b-a) = f[f(f(y))+x]+2x + 2(b-a)

So f(f(x)+y) = f[f(f(y))-x] + 2x + 2(b-a), but comparing this with (1) shows that 2(b-a)=0

So f(a)=f(b)\Longrightarrow a=b and f in injective


proof that f(x)=x

Letting x=y=0 gives f(f(0))=f(f(f(0))), but fince f is injective f(0)=f(f(0))\Longrightarrow f(0)=0

Letting x=0 \Longrightarrow f(y)=f(f(f(y)))\Longrightarrow f(f(y))=y \qquad(2)

Letting y=0 \Longrightarrow f(f(x))=f(-x)+2x \qquad(3)

From (2) and (3), x=f(f(x))=f(-x)+2x \Longrightarrow f(-x)=-x

And \boxed{f(x)=x} is the only solution


PostPosted: Tue Nov 10, 2009 6:49 pm  Back to top 
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ocha
Yang-Mills Theory
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#10
number 4

\left\{\begin{array}{l}a^{2} + c^{2} = 10 \\ b^{2} + 2b(a + c) = 60 \\ \end{array}\right.

Putting the equations together
(b + a + c)^2 - (a + c)^2 = 60 \Longrightarrow b = - (a + c) \pm \sqrt {60 + (a + c)^2}

A = b(a - c) = c^2 - a^2 \pm (c - a)\sqrt {60 + (a + c)^2}


Now set c - a = x, c + a = y and (x + y)^2 + (x - y)^2 = 40 \Longrightarrow x^2 + y^2 = 20

A = x(y \pm \sqrt {60 + y^2})

First consider the the case where the \pm is +
By AM-QM,
\sqrt {y^2} + 4\sqrt {\frac {60 + y^2}{16}} \le \sqrt {5}\sqrt {y^2 + \frac {60 + y^2}{4}} = \frac {5}{2}\sqrt {12 + y^2} \qq...

From (1) we have
A = \frac {5}{2}\sqrt {20 - y^2}\left(y + \sqrt {60 + y^2}\right) \le \frac {5}{2}\sqrt {12 + y^2}\sqrt {20 - y^2}

By AM-GM

\frac {5}{2}\sqrt {12 + y^2}\sqrt {20 - y^2} \le \frac {5}{2}\frac {12 + y^2 + 20 - y^2}{2} = 40

Equality when y = \pm 2

So the Maximum is \boxed{40} equality when (a,b,c) = (1, - 3,10), ( - 1,3, - 10)


Now consider the case where \pm is -
A = x(y - \sqrt {60 + y^2})

But from symmetry we only have to negate our previous solutions

Minimum = \boxed{ - 40} when (a,b,c) = ( - 3,1,10), (3, - 1, - 10)



PostPosted: Wed Nov 11, 2009 1:30 am  Back to top 
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azjps
Yang-Mills Theory
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#11
and number 5
Let x = \text{min}\,(S), and let y \in S with y \neq x. Then z : = \frac {x^2}{y - x} \in S, and z = \frac {x^2}{y - x} \ge x \Longrightarrow y \le 2x,\ \forall y \in S. Hence z = \frac {x^2}{y - x} \le 2x \Longrightarrow y \ge \frac {3x}{2}. Suppose z \neq x; then the same bounds for y apply to z. If z \neq y, wlog y < z; then \frac {3x}{2} \le y < z \le 2x, and 2x = \frac {x^2}{x/2} < \frac {y^2}{z - y} \in S, contradiction. Thus either y = z = \frac {x^2}{y - x} \Longrightarrow y^2 - yx - x^2 = 0 \Longrightarrow (2y - x)^2 = 5x^2, infinite descent gives no solutions, or x = z = \frac {x^2}{y - x} \Longrightarrow y = 2x. Hence the set of solutions is \boxed{\{n,2n\},\, \forall n \in \mathbb{Z}_{ > 0}}, which we easily verify work.


PostPosted: Wed Nov 11, 2009 5:09 pm  Back to top 
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TRAN THAI HUNG
Riemann Hypothesis
Riemann Hypothesis

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#12
Here is the solution for 3
It is easy to prove
\angle KBM = \angle MCB = \angle HMC
then
\begin{array}{l} \Delta BKM \sim \Delta MHC \\ \Rightarrow BK*HC = KM*MH \\ \Rightarrow \Delta KBD \sim \Delta ECH \\ \Righta...
Then MKNB can be inscribed in a circle
\angle KNM = \angle KEM = \angle KBM
Then KMNB can be inscribed in a circle
\begin{array}{l} \angle MNB = \angle AKH \\ \Rightarrow \angle BNC + \angle BAC = 180^0 \\ \end{array}
Can you give any comment about this test?Which is the hardest to you?
@to Ocha how do you have\therefore f(f(x)+a)=f(f(x)+b) ?
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PostPosted: Sat Nov 14, 2009 3:34 am  Back to top 
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