QuoteOriginally posted by: krmy statement is correct but I put 5.7.9 on the wrong list - that one is =3 mod 4, so the lists have the same # of elements, # of reps is 0. Etuka, I don't follow your reasoning. If M = a^2 + b^2, 4M = (2a)^2 + (2b)^2, and 4M = 0 mod 4 - i.e. just because a number is 0 mod 4 doesn't mean it has no representation as a sum of two squares.Sorry, I should read entries more carefully. Of course your statement is correct, though I am not sure what you are asking me about. I thought you were talking about prime divisors for reasons I can't remember. If we are considering prime divisors, the number of representations isThis follows because the function that counts representations is multiplicative, so just consider the prime powers case, then multiply across primes for the result. Powers of 2 have exactly one representation as a sum of powers, and so are transparent as far as the representation count is concerned.

ok, I could've read your text a little closer as well... seemed to sound like "p=3 mod 4 must appear an even # of times, so N=0 mod 4 cannot have rep as a^2 + b^2"...anyhow, yours is the product version of what I wrote with d(n), people here can have some fun with the combinatoricsnow if somebody could just give a similar formula for N = a^2 + 5.b^2...

It is interesting to note that most Highly Composite Numbers (HCNs) that followed by a prime could be uniquely represented in a form a^2+b^2-1.It is easy to state because of well known fact that only and only if a prime number is in a form p=4k+1 it could be represented as a sum of two squares in only one way. This fact was known to Fermat and was proved by Euler.HCN 2520 looks very special as it is a maximum number out of all HCNs that could be represented in a form a^2+b^2-1, where two numbers a,b are consecutive integers and a^2+b^2 is a prime number.2520 = 35^2 + 36^2 - 1There are smaller HCNs like this:180 = 9^2 + 10^2 - 160 = 5^2 + 6^2 - 112 = 2^2 + 3^2 -14 = 1^2 + 2^2 - 1This maximum property of 2520 is a conjecture.Many HCNs could be represented in a form a^2 - 1, where "a" usually a prime number:24 = 5^2 - 1 (= 3^2 + 4^2 - 1)48 = 7^2 - 1120 = 11^2 - 1360 = 19^2 - 1840 = 29^2 - 1 (= 20^2 + 21^2 - 1)1680 = 41^2 - 15040 = 71^2 - 117297280 = 4159^2 - 1It looks like HCN 17297280 is a maximum known HCN that could be represented as a^2-1. Nobody proved that number "a" has to be a prime if a^2-1=HCN. It is a conjecture.

QuoteOriginally posted by: EtukaI can't see how someone could fail to be impressed by the Ramanujan story - it gives the tiniest insight into the quality of his relationship with numbers. There is a vanishingly small number of truly intuitive arithmeticians, let alone of his quality.I just found out today (reading a paper to be published by Mike Hirschhorn, whose web site is quoted by kr below) that Ramanujan's observation re 1729 is an elementary consequence of a deeper observation contained in his Lost Notebook. In my own words, the observation goes as follows:1. We can find non-trivial integral solutions for x^n + y^n = z^n, for n = 2, but not for n > 2, in particular there no solutions for x^3 + y^3 = z^3 (Fermat's Last Theorem, assumed to be true, before the proof).2. Ramanujan's question was: Let's consider n=3. Can we find solutions when x^3 + y^3 = z^3 just fails? More precisely, can we find a triplet of integers {x, y, z}, such that x^3 + y^3 = z^3 + 1? (I think he also asked the question for x^3 + y^3 = z^3 - 1, but I don't have the paper in front of me).3. Ramanujan found the most general answer to the above question: He obtained a recursion relation that any triplet of integers {x_i, y_i, z_i}, for any integral label i > 2, must satisfy, in terms of the preceding 2 triplets. He also obtained the first 2 triplets: {x_1, y_1, z_1} and {x_2, y_2, z_2}, so all the higher ones were determined. The lowest (smallest) non-trivial triplet {x_3, y_3, z_3} = {9, 10, 12}.4. I doubt if Ramanujan left any proof of how he obtained the above result. Herschhorn's proof seems to me to rely on knowing the answer (the recursion relation) and working backwards to find a starting point. I also don't know if Ramanujan investigated similar problems for higher n's, and for larger breakdowns of Fermat's last theorem (by more than 1).5. I find the question as Ramanujan posed it (look for minimal departures of Fermat's last theorem to be really elegant.

Thanks to LongTheta and Ramanujan here is a brainteaser:Find all triplets of positive integers {x,y,z) such that x^3 + y^3 + z^3 = w^3, where w is also an integer number.For example: {3,4,5) and 3^3 + 4^3 + 5^3 = 6^3.Note, the solution of similar problem for integer pairs {x,y} such that x^2 + y^2 = z^2 (where z is also an integer number) was known to Ancient Greeks.PS A new topic started: Cubes.http://wilmott.com/messageview.cfm?cati ... adid=19602