方幂和
定义f(n)为能用两种方式表示成两个不同正整数的n次方幂的和的最小正整数,求f(n)的值 一些已经得到的结果：$f(1)=5=1^1+4^1=2^1+3^1$
$f(2)=65=1^2+8^2=4^2+7^2$
$f(3)=1729=1^3+12^3=9^3+10^3$
$f(4)=635318657=59^4+158^4=133^4+134^4$ 1375298099 = 3^5 + 54^5 + 62^5 = 24^5 + 28^5 + 67^5
160426514 = 3^6 + 19^6 + 22^6 = 10^6 + 15^6 + 23^6 http://oeis.org/search?q=5%2C65%2C1729&language=english&go=Search
Randy Ekl discovered that a number that can be written in two ways as a sum of two fifth powers exceeds 2^74 and one that can be written in two ways as a sum of two sixth powers exceeds 2^89. - Jonathan Vos Post, Nov 28 2007
According to the Mathworld links below, a(5) and a(6), if they exist, exceed 1.02*10^26 and 7.25*10^26, respectively. The page at the SquaresOfCubes link below says Stuart Gascoigne did an exhaustive search and found in Sep 2002 that no a(5) solution less than 3.26*x10^32 exists. My exhaustive search has determined that any solutions for n > 5, if they exist, must exceed 2^96 (about 7.92x10^28). - Jon E. Schoenfield (jonscho(AT)hiwaay.net), Dec 15 2008 不错!很有意思的帖子 Guy, Unsolved Problems In Number Theory, 3rd edition, D1, writes, "... it is not known if there is any nontrivial solution of a5+b5=c5+d5. Dick Lehmer once thought that there might be a solution with a sum of about 25 decimal digits, but a search by Blair Kelly yielded no nontrivial solution with sum ≤1.02×1026."
At F30, Guy writes, "... x5 is a likely answer to the following unsolved problem of Erdos. Find a polynomial P(x) such that all the sums P(a)+P(b) (0≤a<b) are distinct."
The book was published in 2004. I don't know whether there has been any progress since.
http://math.stackexchange.com/questions/226333/generalised-hardy-ramanujan-numbers Hardy-Ramanujan Numbers
Hardy-Ramanujan numbers
http://oeis.org/wiki/Hardy-Ramanujan_numbers
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