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#1
02-26-2003, 12:50 AM
 Frank Bank Member Join Date: Jan 2003 Posts: 179
Maybe this will liven it up around here

This is a pretty neat result from the world of number theory, maybe someone can explain why it works:

Pick a number... you say 74? All right,
how many factors does 74 have? 4 (74,37,2,1)
how many factors does 37 have? 2 (37,1)
how many factors does 2 have? 2 (2,1)
how many factors does 1 have? 1 (1)

Sum of Cubes = 4^3 + 2^3 + 2^3 + 1^3 = 81
Square of Sum = (4+2+2+1)^2 = 81

Try another number... let's say 36 since it has many divisors:
how many factors does 36 have? 9 (36,18,12,9,6,4,3,2,1)
how many factors does 18 have? 6 (18,9,6,3,2,1)
how many factors does 12 have? 6 (12,6,4,3,2,1)
how many factors does 9 have? 3 (9,3,1)
how many factors does 6 have? 4 (6,3,2,1)
how many factors does 4 have? 3 (4,2,1)
how many factors does 3 have? 2 (3,1)
how many factors does 2 have? 2 (2,1)
how many factors does 1 have? 1 (1)

Sum of Cubes = (9^3)+(6^3)+(6^3)+(3^3)+(4^3)+(3^3)+(2^3)+(2^3)+(1 ^3) = 1296
Square of Sum = (9+6+6+3+4+3+2+2+1)^2 = 1296
#2
02-26-2003, 01:51 AM
 Grace Member Join Date: Sep 2002 Posts: 3,105

I protest!!! Twins will be either both sums of cubes or both squares of sums, if identical, but fraternal twins can be a sum of cubes and a square of sums. (So if it works, it's only by accident.)
#3
02-26-2003, 02:02 AM
 Frank Bank Member Join Date: Jan 2003 Posts: 179

Barbie + twins = The Barbie Twins. Now we're cookin' with gas, sister!
#4
02-26-2003, 02:22 AM
 Grace Member Join Date: Sep 2002 Posts: 3,105

The Barbie Twins are coming soon to a thread near you.
#5
02-27-2003, 10:34 AM
 NoName Site Supporter Site Supporter Join Date: Nov 2001 Posts: 5,814

Here is a sketch of a proof:

Take 36. The reason it has 9 divisors is that 36 = 2^2 * 3^2, and any number 2^a * 3^b for a in (0, 1, 2), b in (0, 1, 2) works. The number of divisors for each of those (1+2)*(1+2) = 9 numbers can be determined similarly. It is easy to see that the set of number of factors in each factor can be found by multiplying each member of (1, 2, 3) by each member of (1, 2, 3).

Similarly for any number p1^a1 * p2^a2 * ... * pn^an the analogous set can be found by taking one number from (1, 2, ... a1+1), one from (1, 2, ...a2+1), ... and one from (1, 2, ...an+1), multiplying them all, and cycling over all possibilities.

Presumably we already know that the property holds for any set (1, 2, ... n). So by induction we only need to prove that the property is "multiplicative" in the sense that if

(1) (a1+a2+...an)^2 = (a1^3+a2^3+...+an^3), and
(2) (b1+b2+...bn)^2 = (b1^3+b2^3+...+bn^3), then

(3) (a1b1+a1b2+ ...+a1bn +... +anb1+anb2+ ... anbn)^2 =
([a1b1]^3+[a1b2]^3+ ...+[a1bn]^3 +... +[anb1]^3+[anb2]^3+ ... [anbn]^3)

But this follows directly by multiplying (1) by (2).

QED.

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