Ramanujan
 

Hardy's cab number & Ramanujan

Today is Ramanujan's birth day and I thought I would give a small variation of the above,

As is well known 1979 is the smallest number which can be expressed as the sum of

two cubes of natural numbers in two different ways.

If one of the four Diophantine variables were to belong to Z we get

91 = 3^3 + 4^3 = 6^3 - 5^3.

Q: Is 91 the smallest number that can be expressed in this manner?

A.K. Devaraj

Not 1979
 

[quote=devarajkandadai;154500]Hardy's cab number & Ramanujan

As is well known 1979 is the smallest number which can be expressed as the sum of two cubes of natural numbers in two different ways.

[/quote]

Of course, you mean 1729. I don't believe 1979 is the sum of two cubes, even in only one way:

1979 - 1^3 = 1978 not a cube
1979 - 2^3 = 1971 not a cube
1979 - 3^3 = 1952 not a cube
1979 - 4^3 = 1915 not a cube
1979 - 5^3 = 1854 not a cube
1979 - 6^3 = 1763 not a cube
1979 - 7^3 = 1636 not a cube
1979 - 8^3 = 1467 not a cube
1979 - 9^3 = 1250 not a cube
1979 - 10^3 = 979 not a cube
1979 - 11^3 = 648 not a cube
1979 - 12^3 = 251 not a cube
1979 - 13^3 < 0

Yep, just what I thought :smile:

[QUOTE=devarajkandadai;154500]Hardy's cab number & Ramanujan

Today is Ramanujan's birth day and I thought I would give a small variation of the above,

As is well known 1979 is the smallest number which can be expressed as the sum of

two cubes of natural numbers in two different ways.

If one of the four Diophantine variables were to belong to Z we get

91 = 3^3 + 4^3 = 6^3 - 5^3.

Q: Is 91 the smallest number that can be expressed in this manner?

A.K. Devaraj[/QUOTE]I doubt it, since you mention the set Z then I expect an infinitude of negative values will be smaller.

[QUOTE=NBtarheel_33;154501]Of course, you mean 1729. I don't believe 1979 is the sum of two cubes, even in only one way:

1979 - 1^3 = 1978 not a cube
1979 - 2^3 = 1971 not a cube
1979 - 3^3 = 1952 not a cube
1979 - 4^3 = 1915 not a cube
1979 - 5^3 = 1854 not a cube
1979 - 6^3 = 1763 not a cube
1979 - 7^3 = 1636 not a cube
1979 - 8^3 = 1467 not a cube
1979 - 9^3 = 1250 not a cube
1979 - 10^3 = 979 not a cube
1979 - 11^3 = 648 not a cube
1979 - 12^3 = 251 not a cube
1979 - 13^3 < 0

Yep, just what I thought :smile:[/QUOTE]
Tks; the error was just a typo - the cab number was 1729.
Devaraj

A nice page is [URL]http://euler.free.fr/taxicab.htm[/URL] though unfortunately it does not seem to have been updated for about 2 years.

Your puzzle is defined by the author as cabtaxi(2).

Edit: I just noticed that the new results are now being presented on a new page [URL]http://cboyer.club.fr/Taxicab.htm[/URL]

[quote=devarajkandadai;154525]Tks; the error was just a typo - the cab number was 1729.
Devaraj[/quote]
I don't wish to put a dampener on Hardy's famous anecdote,
but I suspect he might have moulded it so that the punchline
had maximum effect: the interesting property of 1729 raises
gasps of incredulity in the layman.
OTOH knowing that 9^3=729 and 12^3=1728 makes 1729
easy to remember.

David

[quote=devarajkandadai;154500]Hardy's cab number & Ramanujan

Today is Ramanujan's birth day and I thought I would give a small variation of the above,[/quote]

Well, I saw another astonishing quote from Ramanujan in 1913. According to him Factorial(n) + 1 = m^2, then n=4,5,7 only.
I wrote a JAVA program and did as follows:
1. For a given number, calculated the factorial
2. Added 1 to it
3. Calculated the square root of the result
4. Checked if its a perfect square root (ie. sqrt(4) = 2.000000)

Guess what, I tried upto n=22167 and DIDNOT find a perfect m.

It still amazes me, how Ramanujan could state this in 1913?

Eager to know more about Ramanujan's unbelievable conjectures.

-Sastry

spkarra, reading your post I wondered wich roots would have the equation [TEX]\Gamma(x) = x^2[/TEX]

Of course we have [TEX]x_0 = 1[/TEX]
But plotting both functions, it seems there is another intersection near [TEX]x=3,5[/TEX]

Is there any way of getting a closed form of that root?

[QUOTE=Damian;181423]spkarra, reading your post I wondered wich roots would have the equation [TEX]\Gamma(x) = x^2[/TEX]

Of course we have [TEX]x_0 = 1[/TEX]
But plotting both functions, it seems there is another intersection near [TEX]x=3,5[/TEX]

Is there any way of getting a closed form of that root?[/QUOTE]

3.5? Uhm. What? [TEX]\Gamma(x) - x^2 = 0[/TEX] has a root at 5.036722570531942...

Ahh, it looks like you were solving [TEX]\Gamma(x) - x = 0[/TEX], which has a root at 3.562382285390843...

Yes, I was solving those ecuations

[TEX]\Gamma(x) - x^1 = 0[/TEX]
[TEX]\Gamma(x) - x^2 = 0[/TEX]
[TEX]\Gamma(x) - x^3 = 0[/TEX]
[TEX]\Gamma(x) - x^4 = 0[/TEX]
[TEX]\Gamma(x) - x^5 = 0[/TEX]

for [TEX]x > 1[/TEX]

with I tried on wxMaxima:

find_root(x!-x^1, x, 1.01, 100);
find_root(x!-x^2, x, 1.01, 100);
find_root(x!-x^3, x, 1.01, 100);
find_root(x!-x^4, x, 1.01, 100);
find_root(x!-x^5, x, 1.01, 100);

and yielded the following numbers

2.0
3.562382285390898
5.036722570588711
6.464468490129385
7.861923212307213

I was wondering if there were a closed form for these numbers.

[QUOTE=spkarra;181048]Well, I saw another astonishing quote from Ramanujan in 1913. According to him Factorial(n) + 1 = m^2, then n=4,5,7 only.
I wrote a JAVA program and did as follows:
1. For a given number, calculated the factorial
2. Added 1 to it
3. Calculated the square root of the result
4. Checked if its a perfect square root (ie. sqrt(4) = 2.000000)

Guess what, I tried upto n=22167 and DIDNOT find a perfect m.

It still amazes me, how Ramanujan could state this in 1913?

Eager to know more about Ramanujan's unbelievable conjectures.

-Sastry[/QUOTE]Many years ago I verified this conjecture up to n = 100M. A simple algorithm is [b]much[/b] more efficient than yours.

Using only pencil and paper it is straightforward to verify for n up to 1000 or so. Very tedious and requires attention to accuracy, but straightforward.

I won't immediately reveal how to test the conjecture for small n, but I will give a hint: quadratic residues.


Paul