Perfect roots

grandpascorpion · 2006-09-22, 18:03

SOPF(N) =sum of prime factors of N
ND(N) = num of divisors of N
SOD(N) = sum of divisors of N

Let N=35,
SOPF(N)=5+7=12
ND(N)=2*2=4
SOD(N)=35+7+5+1 = 48
SOPF(N)*ND(N)=SOD(N)

35 is the only number to have this property. Can you prove it?
==============================
Let N=22446139
SOPF(N)=31+67+101*107=306
ND(N) = 2^4 = 16
SOD(N) = 23970816 (calculated in PARI)

(SOPF(N)*ND(N))^2 = SOD(N) !

This is the first such number. Are there an infinite number of these cases?

Can you find such a solution where N is not squarefree? I have searched only through 10^8

==============================
Now can you find a number N such that:

(SOPF(N)*ND(N))^3 = SOD(N)

OR prove that such N cannot exist.

Hint: Check numbers with > 4 prime factors (4 factor answers are possible but 5 would be considerably easier). 3 factor answers are not possible.

I haven't found a solution myself.

grandpascorpion · 2006-09-25, 19:52

1305340105703567993091274 = 2*11*355913*497663*553643*605051

SOD(N) = 2^18 * 3^15 * 7^6 * 13^6
SOPF(N) = 3^5 * 7^2 * 13^2
ND(N) = 2^6

So (SOD(N)) = (SOPF(N) * ND(N))^3

Can anyone find a smaller one?

grandpascorpion · 2006-09-27, 15:36

I found 15 other 6-factor answers. The smallest of which is :

11244010994269831858 = 2 * 719 * 5807 * 8747 * 12149 * 12671

SOD(N) = 2^18 * 3^18 * 5^3 * 11^3
SOPF(N) = 3^6 * 5 * 11
ND(N) = 2^6

grandpascorpion · 2006-09-29, 20:09

N = 86,840,656,159,558,885 = 5*41*79*89*107*127*139*167*191

grandpascorpion · 2006-10-01, 13:57

N=14,844,221,560,107,739=31*71*79*109*127*131*179*263

This is likely minimal

2006-09-22, 18:03	#1
grandpascorpion Jan 2005 Transdniestr 503 Posts	Perfect roots SOPF(N) =sum of prime factors of N ND(N) = num of divisors of N SOD(N) = sum of divisors of N Let N=35, SOPF(N)=5+7=12 ND(N)=22=4 SOD(N)=35+7+5+1 = 48 SOPF(N)ND(N)=SOD(N) 35 is the only number to have this property. Can you prove it? ============================== Let N=22446139 SOPF(N)=31+67+101107=306 ND(N) = 2^4 = 16 SOD(N) = 23970816 (calculated in PARI) (SOPF(N)ND(N))^2 = SOD(N) ! This is the first such number. Are there an infinite number of these cases? Can you find such a solution where N is not squarefree? I have searched only through 10^8 ============================== Now can you find a number N such that: (SOPF(N)ND(N))^3 = SOD(N) OR prove that such N cannot exist. Hint: Check numbers with > 4 prime factors (4 factor answers are possible but 5 would be considerably easier). 3 factor answers are not possible. I haven't found a solution myself. Last fiddled with by grandpascorpion on 2006-09-22 at 18:05*

2006-09-25, 19:52	#2
grandpascorpion Jan 2005 Transdniestr 503 Posts	Found a "cube root" 1305340105703567993091274 = 211355913497663553643605051 SOD(N) = 2^18 3^15 * 7^6 * 13^6 SOPF(N) = 3^5 * 7^2 * 13^2 ND(N) = 2^6 So (SOD(N)) = (SOPF(N) * ND(N))^3 Can anyone find a smaller one?

2006-09-27, 15:36	#3
grandpascorpion Jan 2005 Transdniestr 503 Posts	Smaller solution I found 15 other 6-factor answers. The smallest of which is : 11244010994269831858 = 2 * 719 * 5807 * 8747 * 12149 * 12671 SOD(N) = 2^18 * 3^18 * 5^3 * 11^3 SOPF(N) = 3^6 * 5 * 11 ND(N) = 2^6

2006-09-29, 20:09	#4
grandpascorpion Jan 2005 Transdniestr 503 Posts	9-factor solution N = 86,840,656,159,558,885 = 5417989107127139167191 Last fiddled with by grandpascorpion on 2006-09-29 at 20:47 Reason: Better solution

2006-10-01, 13:57	#5
grandpascorpion Jan 2005 Transdniestr 767₈ Posts	Smaller 8-factor "cube" root N=14,844,221,560,107,739=317179109127131179263 This is likely minimal Last fiddled with by grandpascorpion on 2006-10-01 at 13:58*

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Extracting k-th roots mod m	Dubslow	Miscellaneous Math	8	2012-12-13 21:35
Roots of 1 mod 2^n	fenderbender	Miscellaneous Math	17	2010-11-16 16:25
primtive roots mod p^2	Peter Hackman	Math	2	2007-10-24 06:41
What is the roots of NFS algorithm?	Annunaki	NFSNET Discussion	7	2003-07-29 16:01
Identifing perfect squares and calculating square roots..	dsouza123	Math	2	2003-07-19 17:17