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Erdos
  
Number of games in database: 1
Years covered: 1922


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 page 1 of 1; one game  PGN Download 
Game  ResultMoves YearEvent/LocaleOpening
1. Erdos vs Lichtner 1-0181922ViennaD00 Queen's Pawn Game

Kibitzer's Corner
Oct-22-06  Milo: Obviously this is not the great Hungarian mathematician Paul Erdos. But we should discuss mathematics here anyways, maybe. To begin, a problem from some national olympiad:

Does there exist a function f, not identically 0, with the property that, for all a,b in the open interval (0,1), and all real x not equal to zero, f(x) = af(ax) + bf(bx) ?

Mar-16-07  imatos: Seems to me that it doesn't, although I'm probably missing something, since I'd expect olympiad-level problems to be a bit tougher than this. :-)

Since f(x) is defined for all real x except zero, we have as a special case for x = 1 and b = a:

f(1) = 2af(a),

and thus f(a) = f(1)/(2a) for all a in (0,1), and thus for all x in (0,1), f(x) = f(1)/(2x).

But then for any 0 < x < 1:

f(x) = 2af(ax) = 2af(1)/(2ax) = f(1)/x,

which contradicts the above. Q.E.D.

Mar-16-07  azaris: <Since f(x) is defined for all real x except zero, we have as a special case for x = 1 and b = a:

f(1) = 2af(a),

and thus f(a) = f(1)/(2a) for all a in (0,1), and thus for all x in (0,1), f(x) = f(1)/(2x).

But then for any 0 < x < 1:

f(x) = 2af(ax) = 2af(1)/(2ax) = f(1)/x,

which contradicts the above.>

Not really, it just shows that f(1) = 0. The function f could still be nonvanishing outside (0,1].

Mar-16-07  imatos: Well, if the function is vanishing inside (0,1] and f(x) = 2af(ax) for 0 < a < 1, then for any x > 1, just take some arbitrary 0 < a < 1/x and then:

f(x) = 2af(ax) = 0,

since 0 < ax < 1 and thus f(ax) = 0.

Mar-19-07  azaris: OK, here's a simple one:

Is every natural number a difference of two perfect squares?

Mar-19-07
Premium Chessgames Member
  WannaBe: <azaris> Do you mean:

0 = x^2 - x^2
1 = 1^2 - 0^2
2 = ??

Mar-19-07  azaris: <WannaBe> Uh-huh.
Mar-19-07
Premium Chessgames Member
  WannaBe: http://news.yahoo.com/s/livescience...
Mar-19-07  azaris: Wow, 248 dimensions. I wonder what they'd think when they hear most mathematicians work with infinite dimensional objects every day and that they're competely mundane. But otherwise a pretty good mainstream article about mathematics.
Jul-13-09
Premium Chessgames Member
  johnlspouge: < <Milo> wrote:

Does there exist a function f, not identically 0, with the property that, for all a,b in the open interval (0,1), and all real x not equal to zero, f(x) = af(ax) + bf(bx) ? >

Answer: No, the only function with the property described is identically 0.

Proof. For any 0 < y < x < 1, set a = b = y / x. Then, algebra yields

y f(y) = x f(x) / 2.

For any 0 < y < 1, pick a fixed x in (y, 1). For any natural number n, pick a sequence y < x [1] < ... < x [n - 1] < x [n] = x. Then,

y f(y) = x f(x) / pow (2, n).

Take limits as n tends to infinity, to show y f(y) = 0, so f(y) = 0 for all 0 < y < 1.

Some kibitz :)

Jul-13-09
Premium Chessgames Member
  johnlspouge: I missed the condition that x is not restricted to (0,1), but the proof obviously applies to all real numbers.
Aug-09-09  Eduardo Leon: <azaris> For a number to be the difference of two perfect squares, there's a simple test: divide it by 4 and take the remainder. It's the difference of two perfect squares if and only if the remainder is not 2.

Proof:

Take two consecutive perfect squares, say n^2 and (n+1)^2. The difference between them is 2n+1, which must be an odd number and could be any odd number. This proves any odd number could be the difference between perfect squares.

The only way that an even number can be the difference between two perfect squares is that these two squares are either both odd or both even. Take two squares such that the preceding assertion is true, say, n^2 and (n+2m)^2. The difference between them is 4m(m+n), which must be multiple of 4 and could be any multiple of 4.

So we can make the following assertion: Any odd number is the difference between two perfect squares, and an even number is the difference between two perfect squares if and only if it's a multiple of 4.

We can rephrase the last assertion: A number is the difference between two perfect squares if and only if the number, when divided by 4, doesn't yield 2 as the remainder.

Jan-11-10
Premium Chessgames Member
  WannaBe: Okay, okay...

So, I see this name, Paul Erös pop up on this site a while back, and also a name of a book, 'The Man Who Only Loved Numbers'. Lo and behold, guess what I got for Christmas.

What a book! (And I am only 1/2 done!) Did not know, so many of the people in the book that I have come across! (Okay, okay, so it's only one) Dr. Prof. Bruce Rothchild.

When I was at UCLA, I never had the chance to take a class with Dr. Rothchild, I now look back with regret, if I had heard of Erdös, and have read this book, I would have asked Dr. Rothchild so many questions!!

I guess, if that I had been smart enough to study hard, and advance my knowledge of math, I <<may>> have ended up writing a paper with Dr. Rothchild. And get an Erös Number myself! Alas, it was not written in the stars.

Then again, it's about the same odd as me being named the MVP of the World Series. Since I got stuck, and couldn't get past basic junior level Abstract Algebra. And had to switch from Math major to Comp. Sci...

May-12-19
Premium Chessgames Member
  Jean Defuse: ...

Probably the Hungarian Austrian problem composer <Géza Erdős>. He settled in 1911 in Vienna, where he lived until the end of his life.

Some examples of his problems:

http://sakk-magazin.hu/4si/pgn/erdo...

And another game of Erdos:

[Event "Vienna"]
[Date "1922.??.??"]
[White "Rund"]
[Black "Erdos"]
[Result "0-1"]
[ECO "C50"]

1. e4 e5 2. Nf3 Nc6 3. Bc4 Bc5 4. d3 d6 5. O-O Nf6 6. Ng5 O-O 7. Be3 Bg4 8. Qd2 Nd4 9. c3 Ne2+ 10. Kh1 Nh5 11. Bxc5 Nhf4 12. Nxf7 Qh4 13. Nxd6+ Kh8 14. Nf7+ Rxf7 15. Bxf7 Qh3 16. Rg1 Bf3 0-1

More information: http://sakk-magazin.hu/4si/kerdes.p...

...

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