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 April 17th, 2012, 09:12 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory property of numbers of factors of factors of a number Ok, now im going to say about a marvelous problem of number theory. Let there be a number,k,such that K = p*q*r*..... where p,q,r..... are prime numbers. Then the factors of K are, {1,p,q,r,......,p*q,p*r,q*r,.......,p*q*r,......p* q*r*.....} Now consider about the number of factors of every factor of K. number of factors of 1 is 1. Number of factors of p,q,r,.......... is 2. Number of factors of p*q is 4 (i.e 1, p, q, pq). The problem is to prove that : If a,b,c,... are respectively the number of factors of every factor of K then, $a^3+b^3+c^3+.....= (a+b+c+.....)^2$ I can prove it. But surprisingly, it is also true if some of the p,q,r,..... are not primes. For this one, i am totaly lost. when K = $p^n$ this theorem will generate the formula 1^3+2^3+.....=(1+2+....)^2 as a special case.
April 17th, 2012, 09:48 AM   #2
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Re: property of numbers of factors of factors of a number

Quote:
 Originally Posted by mathbalarka If a,b,c,... are respectively the number of factors of every factor of K then, $a^3+b^3+c^3+.....= (a+b+c+.....)^2$
Perhaps I misunderstand you, but if K = 2 than the factors of K are 1 and 2 which have 1 and 2 factors, respectively. But 1^3 + 2^3 is not equal to (1+2)^2.

April 17th, 2012, 10:24 AM   #3
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Re: property of numbers of factors of factors of a number

Quote:
 Originally Posted by CRGreathouse ...But 1^3 + 2^3 is not equal to (1+2)^2.
No?

Don't we have:

$\sum_{k=0}^n$$k^3$$=$$\sum_{k=0}^n\(k$$\)^2=$$\fra c{n(n+1)}{2}$$^2$ ?

:P

Maybe you meant:

$1^3+3^3\ne(1+3)^2$

 April 17th, 2012, 10:33 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: property of numbers of factors of factors of a number Oops. I was cubing rather than squaring the RHS.
April 17th, 2012, 09:36 PM   #5
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Re: property of numbers of factors of factors of a number

Quote:
 Originally Posted by MarkFL maybe you meant $1^3+3^3=/= (1+3)^2$
Which is impossible. A number cannot have such factor which have 1, 3 factors respectively

 April 19th, 2012, 09:04 PM #6 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: property of numbers of factors of factors of a number Hey guys, i think i found a marvelous application of this theorem. consider the diophantine equation : $y^2=x^3+17$ Then consider the relationship : $(a+b+c+.....)^2= a^3+b^3+c^3+......$ where a,b,... are the factors of factors of K. Comparing both relations, we conclude that $y= a+b+c+.....$ and $x^3+17= a^3+b^3+c^3+.....$ By random interation, we see that 17 = 1^3 + 2^3 + 2^3 Now, x^3 cannot be equal to sum of two cubes (by FLT for n=3). Then the factors of factors of K are 1, 2, 2 and x. Then the factors of K are 1, p, q, t. Then t must be pq and so, x is 4 and y is 9 or -9. Thats how this theorem finds a solution to this diophantine equation.
 April 20th, 2012, 05:29 AM #7 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: property of numbers of factors of factors of a number Ok, i v proved this theorem if K = (p^m)(q^n) now our goal is to prove it for any such K which is the product of distinct powers of non equal primes.It will be a hard work so if anyone helps me, i will be really very greatful to him.
 April 20th, 2012, 11:54 AM #8 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: property of numbers of factors of factors of a number Ok then, let us consider about K such that K = p*q*r*..... where p,q,r,..... are primes.( note: the numbers of the prime factors; p,q,r,... are n) Theorem: If the number of factors of factors of K are a,b,c,.... respectively; then it satisfies the relationship: $a^3+b^3+c^3+.....= (a+b+c+.....)^2$ Proof: the number of factors of factors of K are: {1,2,2,2,2,......2,4,4,4,4,4,........4,8,8,8,8,... ...,8,......} We can show that the distinct numbers which the factors of factors of K(i.e 2, 4, 8, 16 e.t.c) are all of the form: 2^n We have done the first part (general form of the factors) now all we have to do is to consider about the second part (how many factors are there) the numbers of 1 digit factors of K (i.e p,q,r,...) is n or $\binom{n}{1}$ the numbers of 2 digit factors of K (i.e pq,qr,...) is n(n-1) $\binom{n}{2}$ Now we can easily show that the numbers of p digit factors of K is $\binom{n}{p}$ Now the main proof: Then the summation of cube of the number of factors of factors of K is: $\binom{n}{0} (2^0)^3 + \binom{n}{1} (2^1)^3 + \binom{n}{1} (2^2)^3 + ...... + \binom{n}{n} (2^n)^3$ by newtons binomial formula, we conclude that the formula above is equivalent to (1+ 2^3)^n which is equals to 9^n. On the other hand, the square of summation of the number of factors of factors of K is: $(\binom{n}{0} 2^0 + \binom{n}{1} 2^1 + \binom{n}{2} 2^2 + ...... + \binom{n}{n} 2^n)^2$ which is equals to (1+2)^(2n) = 9^n. Q.E.D
 April 20th, 2012, 12:48 PM #9 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: property of numbers of factors of factors of a number I will show you the proof of this theorem for K=(P^m)(Q^n) after i get some sleep.
 April 20th, 2012, 09:38 PM #10 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: property of numbers of factors of factors of a number Theorem: If K=$p^n q^m$(p,q are primes) and if the numbers of factors of factors of K are a,b,c,... respectively, then then it satisfies the relation : $a^3+b^3+c^3+.....= (a+b+c+.....)^2$ Proof: factors of factors of K are {1,p,p^2,.....,p^n,q,q^2,......,q^n,pq,p^2q,...... } Now we will make an array of the number of factors of factors of K. The line numbers will be denoted by 1,2, etc. To be continued on the next post.

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