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March 13

Formula for magic triples??

We know that {2, 2} is a magic pair because 2 + 2 = 4 and 2 * 2 = 4. This is the only magic pair whose elements are both natural numbers. However, there are infinitely many magic pairs in the real numbers, including:

{3, 1 1/2} {4, 1 1/3} {5, 1 1/4}

...and so on. In fact, there is a magic tuple for every nonzero x whose formula is {1+x, 1+(1/x)}. Setting x to 1 in the above formula gives the familiar {2, 2}.

How about magic triples?? I know of only one magic triple, {1, 2, 3}. Is there a formula for magic triples in the real numbers?? Georgia guy (talk) 14:17, 13 March 2016 (UTC)[reply]

Assuming a magic triple is a triple (A,B,C) such that A+B+C=ABC, then for any fixed A,B such that

AB\neq 1

, this defines a linear equation for C whose solution is

C=(A+B)/(AB-1)

. So there is a two-parameter set of solutions. Sławomir
Biały 14:52, 13 March 2016 (UTC)[reply]

It's not hard to show that {a, b, c} = {1, 2, 3} is the only solution in positive integers, and that the only other solutions in integers are {a, b, c} = {-1, -2, -3} and {a, b, c} = {n, -n, 0}. Also (4, 2, 1, 1) is a magic quadruple, (5, 2, 1, 1, 1) is a magic pentuple, and so on. Other magic combinations in positive integers are possible, e.g. (4, 3, 1, 1, 1, 1, 1), (2, 2, 2, 1, 1). --RDBury (talk) 19:22, 13 March 2016 (UTC)[reply]

By the way, the magic pairs problem appears at the bottom of a PDF file on odd abundant numbers, and complementary Beatty sequences as in Rayleigh's theorem also give magic pairs. GeoffreyT2000 (talk) 22:37, 13 March 2016 (UTC)[reply]

Restricted Waring problem

We know the solution to the Waring problem; for any positive integer n, there is an integer k such that every positive integer is a sum of at most k powers of positive integers. However, I cannot figure out how to search the literature for the answer to the question: For any positive integer n, there is an integer k such that all sufficiently large positive integers are the sum of k non-zero powers of positive integers. I'm particular interested in n = 2, but .... — Arthur Rubin (talk) 17:21, 13 March 2016 (UTC)[reply]

Never mind; I figured it out. By the four square theorem, every number >= 1 is a sum of 1 through 4 squares. As 5² = 3² + 4², every number >= 26 is a sum of 3 through 5 squares, every number >= 51 is a sum of 5 or 6 squares, and every number >= 76 is a sum of 7 squares. — Arthur Rubin (talk) 17:41, 13 March 2016 (UTC)[reply]

If anyone is reading this, as 3² = 2² + 2² + 1², it follows that any integer > 34 is a sum of 6 positive squares. Investigation of intermediate steps shows that any integer > 19 is a sum of 6 positive squares. Experimental evidence suggests that the largest number not the sum of 5 positive squares is 33, but

7\times 2^{2n+1}

is not the sum of 4 positive squares. — Arthur Rubin (talk) 01:31, 15 March 2016 (UTC)[reply]

169 = 13² = 12² + 5² = 12² + 4² + 3² = 1² + 2² + 8² + 10². Since every natural number is the sum of at most four positive squares, every integer greater or equal to 169 is the sum of five positive squares. You've already checked the small cases, so this proves that every integer greater than 33 is the sum of five positive squares. – b_jonas 16:36, 19 March 2016 (UTC)[reply]

Connected rings

Does there exist a connected ring R (a nontrivial commutative ring with no idempotents other than 0 and 1) such that the contravariant functor Hom (–, R) on the category of commutative rings sends arbitrary small products to coproducts? Note that this is always true for finite products and arbitrary connected rings. GeoffreyT2000 (talk) 22:56, 13 March 2016 (UTC)[reply]