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At 10:43am on March 1, 2016, dasari naga vijay krishna said…

sorry sir,

At 6:42pm on January 26, 2015, Devarakonda said…

When They Resisted The Booming
And Shelling Of The Colonial Guns,
Our Founder Fathers Wanted Nothing
But Sovereignty For This Nation,
Let Us Always Defend This Gift
From Our Predecessors.

Happy Republic Day.

At 12:15pm on January 11, 2013, trantancuong said…

Hi Monsieur Jean Yves Rolli.

I am so happy can talking to you continue.

Happy new year.

Pierre De Fermat 's last Theorem.
The conditions:
x,y,z,n are the integers and >0. n>2.
Proof:
z^n=/x^n+y^n.

We have;
z^3=[z(z+1)/2]^2-[(z-1)z/2]^2
Example;
5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125
And
z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2
Example;
5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189
And
z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2
Example
5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216
And
z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2
Example
5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224
General:
z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2

We have;
z^3=z^3+(z-m-1)^3 - (z-m-1)^3.
Because:
z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3]
So
z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3.
Similar:
z^3=z^3+(z-m-2)^3 - (z-m-2)^3.
So
z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3.
....
....
Suppose:
z^n=x^n+y^n
So
z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3.
So
z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3}
Similar:
z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-

At 12:14pm on January 11, 2013, trantancuong said…

Hi Monsieur Jean Yves Rolli. I am so happy can talking to you continue. Happy new year. Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.

At 8:24pm on July 8, 2012, MT Ansari said…

Hi Sir 

      We are working here with elementary classes and developing content for children. Could you suggest us what approach we have to adopt for content in math. 

        

At 9:09pm on November 11, 2010, Devarakonda said…
Advance Eid Mubarak.........!
At 9:08pm on November 11, 2010, Devarakonda said…

At 12:56am on July 10, 2010, N.Balasubramanian said…
It is disturbing to learn that we may have to close. I wish we are able to keep it going at some cost - But what cost, can be discussed if a concrete proposal is made.
At 4:09pm on May 13, 2010, G.S.Shravankumar said…
Dear sir
if i have any queries can i mail you ?
At 6:00pm on April 2, 2010, Manidipa Pal said…
thank you sir for accepting my friend request ...
At 10:20am on March 9, 2010, Manmohan singh said…
Good Morning Sir , i had attended MTTS 2001 and progrmme by casio at DPS RK Puram New Delhi
, I m Directing Mathemagicians club for students and want to organise a workshop for students and teachers , Guide me about the full concept , time , money, person etc
Thanking u
Manmohan Singh
At 11:31pm on March 8, 2010, Manidipa Pal said…
ya..I have seen this video sir,it is very good...background music is awesome,,
At 7:02pm on December 1, 2009, Vinod Kumar P said…
As told by Dr. A. Vijayakumar ( CUSAT) I have sent 10 Tatva CDs and 10 Charitha CDs to you today (1-12-2009)

An appreciation of Tatva
http://www.springerlink.com/content/n85q6j565715v236/

The tatva of mathematics
http://www.hindu.com/edu/2005/01/31/stories/2005013104210600.htm

A historical offering from a maths club
http://www.hinduonnet.com/edu/2006/05/02/stories/2006050200070400.htm

We wish TIME 2009 all success.
With regards,

Vinod Kumar P
Dept. of Mathematics
Payyanur College
pvinodkumar@gmail.com
At 1:36pm on April 25, 2009, Saumil Shrivastava said…
Dear sir, You may not remember me but you taught me mathematics course while I was in IIT Bombay. Its great to see you hear sir. :)
At 5:30pm on March 28, 2009, Anupam Yadav said…
Hello Sir, Is there be any concept of Uniform Continuty in the Topological Spaces?
At 8:38pm on March 17, 2009, Seema Mishra said…
Dear Sir, many thanks for the math-network. At present I'm teaching in a govt. college in Bhopal. I have interest in fuzzy topology, I want to work for D.Sc. now. Could you please help me in this regard , also guide if any fellowship is available for postdoc, Regards
At 2:45pm on March 2, 2009, hareesh kumar k k said…
i appreciate this endeavor.i'm interested in maths teaching and writing for students.i also want 2 participate in TIME2009 and if possible present
a paper on mathsteaching
At 1:10pm on January 6, 2009, avani shah said…
i'm glad to be the member of this group. i will soon start some discussions on my favourite toics on math
At 11:48pm on January 3, 2009, tiju said…
dear sir,
thank u so much 4 this..... its really a great thing......
At 7:08pm on October 10, 2008, Ramanujam said…
Dear sir,
Is there any probleb in that geogebra file?,can you use it?

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