# a problem of mu , pi , epsilon (journal)

Let $p_n$ denotes the $n$ th prime and $\pi(n)$denotes the numbers of primes $\leq$ $n$ . Prove that for $n$ $\geq$ $6$ the $inequality$ holds : $\pi(\sqrt{\displaystyle{\prod_{i=1}^{n}}}p_i)$ $>$ $2n$ ($mu$ $pi$ $epsilon$ $journal$)

Solution :Processing by induction for $n = 6$ ,we have $\pi(\sqrt{\displaystyle{\prod_{i=1}^{6}}}p_i)$ = $\pi(\sqrt 30030)\geq \pi(50)$ = 15 $>$ 6$\cdot$2.

So , this is valid for n = 6.
Now suppose $\pi(\sqrt{p_1p_2..........p_n)} > 2n$ for some n$\geq$ 6.
We will show that $\pi(\sqrt{p_1p_2..........p_np_{n+1})} > 2(n+1) = 2n+2$. That is ,
$\pi(\sqrt{p_1p_2..........p_n}\cdot{\sqrt{p_{n+1}}}$) $> 2n +2$
Since n$\geq$ 6 , then $p_{n+1} \geq p_7 = 17$ and $\sqrt{p_{n+1}} > 4$.
Now , it is easy to say that $\pi(\sqrt{p_1p_2..........p_n}\cdot{\sqrt{p_{n+1}}}$) $\geq$ $\pi(4\sqrt{p_1p_2..........p_n}$

Now, $Bertrand's Postulate$ guarantees a prime number between $k$ and $2k$ for all $k > 1.$ Thus $\pi((4k)-\pi(2k)$)and $\pi((2k)-\pi(k)$ )are each atleast $1$ , which follows that $\pi((4k)-\pi(2k)$)+ $\pi((2k)-\pi(k)$ )$\geq1+1$.
Now observe that ,$\pi(4k)$ = $\pi((4k)-\pi(2k)$)+ $\pi((2k)-\pi(k)$ )+$\pi(k)$.
Then
$\pi(\sqrt{p_1p_2..........p_np_{n+1})}$ $\geq$ $\pi(4\sqrt{p_1p_2..........p_n}$ =
$\pi(4\sqrt{p_1p_2..........p_n}$$\pi(2\sqrt{p_1p_2..........p_n}$ + $\pi(2\sqrt{p_1p_2..........p_n})$)- $\pi(\sqrt{p_1p_2..........p_n}$)$>1+1+2n$ ( since ,$\pi(\sqrt{p_1p_2..........p_n)} > 2n$)//That is $\pi(\sqrt{p_1p_2..........p_np_{n+1})}$ $>$ $2n+2$.
So, we are done .

# the ring of numbers

Elementary number theory is largely about the ring of integers, denoted by the symbol . The integers are an example of an algebraic structure called an integral domain. This means that satisfies the following axioms:

(a) has operations + (addition) and (multiplication). It is closed under these operations, in that if , then and .

(b) Addition is associative: If , then

(c) There is an additive identity : For all ,

(d) Every element has an additive inverse: If , there is an element such that

(e) Addition is commutative: If , then

(f) Multiplication is associative: If , then

(g) There is an multiplicative identity : For all ,

(h) Multiplication is commutative: If , then

(i) The Distributive Laws hold: If , then

(j) There are no zero divisors: If and , then either or .

# Polynomial (us amo)

Suppose that P(x) is a polynomial of degree n such that P(k) = $\frac{k}{k+1}$ for k = 0, 1, 2, …, n . Find the value of $P(n+1)$.

Solution :

Consider an auxiliary polynomial g(x) = (x+1)P(x) – x . g(x) is an $n+1$ degree polynomial (as P(x) is n degree and we multiply (x+1) with it). We note that g(0) = g(1) = … = g(n) = 0 (as the given condition allows (k+1) P(k) – k = 0 for all k from 0 to n). Hence 0, 1, 2, … , n are the n+1 roots of g(x).

Therefore we may write g(x) = (x+1)P(x) – x = C(x)(x-1)(x-2)…(x-n) where C is a constant. Put x = -1. We get g(-1) = (-1+1)P(-1) – (-1) = C(-1)(-1-1)(-1-2)…(-1-n).

Thus 1 = C $(-1)^{n+1}$ (n+1)! gives us the value of C. We put the value of C in the equation (x+1)P(x) – x = C(x)(x-1)(x-2)…(x-n) and replace x by n+1 to get the value of P(n+1).

(n+2)P(n+1) – (n+1) = $(-1)^{(n+1)}$ {(n+1)!} (n+1)(n)(n-1) … (1) implying P(n+1) = ${(-1)^{(n+1)} + (n+1)} {(n+2)}$

# B.stat -2005 – geometry problem

$B.stat-2005$– (Solution)
[by Neelanjan Mondal] At first use the $AM-HM$ inequality ,
$\frac{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}{3}$ $\geq$ $\frac{3}{\alpha+\beta+\gamma}$, note that $\alpha$ be equal to $180^\circ$$P$,
$\beta$ must be equal to $90^\circ$+$P/2$ and $\gamma$ must be equal to $2P$. , that is for any acute angled triangle $PQR$ , put the values of $\alpha ,\beta, \gamma$ , so we obtain ,
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ $\geq$ $\frac{9}{\alpha+\beta+\gamma}$;
=$\frac{9}{270^\circ + \frac{P}{2}+P}$ =$\frac{9}{270^\circ + \frac{3P}{2}}$
= $\frac{18}{540+3P}$.
It is given that $\angle P$ $<$ $90^\circ$ that is $3P + 540^\circ$ $<$ $270^\circ +540^\circ$ =$810^\circ$
that is $\frac{18}{3P + 540^\circ}$ $>$ $\frac{18}{810}$ = $\frac{1}{45}$.
So ,it is easy to say ,$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ $>$$\frac{1}{45}$.
So , we are done .
[Please note that , $1 = \alpha,2 = \beta , 3 = \gamma$]

Here , we have used a lemma (about the incircle , incenter…etc.) that is …

Note that $A+B+C$ = $\pi$ for any acute angled triangle $\Delta ABC$ , dividing by $2$ to the both sides , we get $\frac{A+B+C}{2} = \frac{\pi}{2}$ .
Now , $\angle BIC + \frac{B}{2} + \frac{C}{2}$ = $\pi$ for the triangle $\Delta BIC$.
Now we will express $\angle BIC$ by the terms of $A$.
so ,$\angle BIC$ = $\pi$ – ($\frac{B+C}{2})$ = $\pi - (\frac{\pi}{2} - \frac{A}{2})$.That is equal to $\frac{\pi}{2} +\frac{A}{2}$ = $90^\circ + \frac{A}{2}$.

# solution of regular polygons of problem no . 2

From the given relation , we have $A_1A_2\cdot A_1A_3 + A_1A_2\cdot A_1A_4 = A_1A_3 \cdot A_1A_4$…………..(1)
Also in the cyclic quadrilateral $A_1A_3A_4A_5$ , we have , by the $Ptolemy's Theorem$ ,
$A_4A_5\cdot A_1A_3$ +$A_3A_4\cdot A_1A_5$ = $A_3A_5\cdot A_1A_4$ .
Since $A_1A_2A_3.........................A_n$ is a regular polygon , so we have $A_iA_j = A_jA_k$ for $1\leq i\leq j\leq k\leq n$.
………..(2)
Comparing (1) and (2) , we have $A-1A_4 = A_1A_5$ .
Since the two diagonals follows the relation $A-1A_4 = A_1A_5$ , then it is easy to say that there must be the same number pf vertices between $_1$ and $A_4$ as between $A_1$ and $A_5$.
So , only possibility is $n = 7$
Attachments:

# Regular polygons

In this session , the problems consists the uses of symmectries of the polygon . Here , we illustrate this with the following fasciating facts of the construction of the regular petagons .Of cource there exists a classical ruler and compass construction , but there is an easierway to do it . At first , make the simplest knot, the trefoil knot, on a ribbon of paper , then flattern it as the following knot .After cutting of the two ends of the ribbon , one obtains a regular pentagons .
Problems :
1.Let $A_1A_2....................A_n$ be a regular polygons inscribed in the circle of the centre $O$ and the radius $R$. On the half line $OA_1$ cnoose $P$ such that $A_1$ is between $O$ and $P$ .Prove that
$\displaystyle{\prod_{i=1}^{n}}PA_i$ = $PO^n-R^n$.
2.Let $A_1...........A_7$ be a regular heptagon .Prove that
$\frac{1}{A_1A_2} = \frac{1}{A_1A_3} + \frac{1}{A_1A_4}.$
If the above relation holds then prove that the polygon is a heptagon , that is $n = 7$( $INMO-1992$).
Attachments:

# Problems for 2016

1.Let a , b , c be the lenghts of the edges of a right parallelepiped and d be the the lenght og its diagonal .Prove that $(ab)^2+(bc)^2+(ca)^2 \geq abcd\sqrt3$ .
2.Prove that for 1$\leq x_k\leq 2$ , k = 1………….n the inequality
$\displaystyle{(\sum_{k=1}^{n}}x_k)$$\displaystyle{(\sum_{k=1}^{n}}\frac{1}{x_k})^2$$\leq n^3.$
3.Prove that for every integer n $\geq$ 2 , the number $2^{2^n-2}+1$ is not a prime .
4.Show that there exist infinetely many $a$ such that for any $n$ the number $n^4 + a$ is not an prime .
5.Prove that for any integer n $\geq$ 1 , the number $n^12 + 64$ can be written as the product of four distinct integers .
6.$Hermit$$Identity$ : Prove that the function $f:\mathbb{R}$$\rightarrow \mathbb{N}$ ,

where f(x) = $\lfloor x \rfloor +$$\lfloor{x+\frac{1}{n}}\rfloor+$………+$\lfloor{x+\frac{n-1}{n}}\rfloor$$\lfloor{nx}\rfloor$.
One can easily check that f(x) = 0 for x$\in$ $\big{[0,\frac{1}{n})}$ .
prove that $f$ is a periodic function with period $\frac{1}{n}$
.

7.If a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is not an one -to -one function and there exist a function $g:\mathbb{R}\times\mathbb{R}$ such that $f(x+y) = g(f(x) , y)$ for all x,y $\in$ $\mathbb{R}$.Show that f is a periodic function .
8.Let $f:\mathbb{R}\rightarrow\mathbb{R}$ is a $periodic$ function such that the set {$f(n)|n\in \mathbb{N}$} has infinetely many elements .Prove that the period off $f$ is irrational .

# number theory 2013

Let N be an integer such that N(N – 101) is a perfect square of a positive integer .

Solution :We will not consider the cases where N = 0 or N = 101)
$( N(N-101) = m^2 )$

that is $( N^2 – 101N – m^2 = 0$)

Roots of this quadratic in N is
that is $( \frac{101 \pm \sqrt { 101^2 + 4m^2}}{2}$)

The discriminant must be square of an odd number in order to have integer values for N.

Thus $( 101^2 + 4m^2 = (2k + 1)^2$)
that is $( 101^2 = (2k +1)^2 – 4m^2$)
that is $( 101^2 = (2k +2m + 1)(2k – 2m + 1)$)

Note that 101 is a prime number

Hence we have two possibilities

Case 1:

$( 2k + 2m + 1 = 101^2; 2k – 2m + 1 = 1$)
Subtracting this pair of equations we get $$$4m = 101^2 – 1$$$ or $$$4m = 100 \times 102$$$ or m = 50 $\cdot 51$

This gives N = 2601

Case 2:

$(2k + 2m + 1 = 101 ; 2k – 2m + 1 = 101$) which gives m = 0 or N = 101. This solution we ignore as it makes N(N- 101) = 0 (a non positive square).

Hence the only solution is N = 2601 and there are no other values of N for which N(N-101) is not a perfect square .

# putnam mathematical competitions

Find all f(x) such that $\int[f(x)]^n = [\int f(x)]^n$

# rouche’s theorem

Theorem :Let f and g be two polynomials function on inside a simple closed curve ( e.g a circle ) .Suppose that |f(z)| > |g(z)| for all points  on the curve . Then f and f-g have the number of zeroes ( counting multiples ) interior to the curve .