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 April 3rd, 2009, 08:18 PM #1 Senior Member   Joined: Nov 2007 Posts: 633 Thanks: 0 Diophantine equation : factorial and powers Can we express n! as sum of powered numbers? n! = (a^b + c^d + e^f ....) - (a'^b' + c'^d' + e'^f' ....) where a,a',b,b',c,c' ..... are integers > 1 I'm not looking for a trivial solution. The number of the variable used has to be minimal of course. If we can find a general solution then the factorization problem will be over. April 3rd, 2009, 10:04 PM #2 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: Diophantine equation : factorial and powers Every number can be so represented, factorials are no different from the others. The number of terms will be at least two. 2! = 2^3 - 2^2 3! = 13^3 - 3^7 - 2^2 4! = 2^10 - 10^3 5! = 31^2 - 29^2 6! = 181^2 - 179^2 7! = 1261^2 - 1259^2 8! = 5042^2 - 5038^2 9! = 610^2 - 96^2 - 2^2 10! = 78^4 - 322^3 - 2^3 11! = 6802^2 - 2520^2 - 2^2 12! = 6850^2 + 756^3 - 46^2 The code's bad, but if anyone wants to glance at it for some reason here it is. You could use it as a starting place if you like, but starting over may be better. Code: pw=[];for(n=2,1e4,for(k=2,log(1e8)/log(n),pw=concat(pw,n^k)));pw=vecsort(pw,,8); lf(t,k)=for(i=k+1,#pw,if(pw[i]-pw[i-k]==t,return([pw[i],pw[i-k]])));0 lookfor(t)=for(k=1,(t>>20)+50,l=lf(t,k);if(l!=0,return(l))) nice(v)=my(a,b,c,d);if(v==0,print("?");return());b=ispower(v,,&a);d=ispower(v,,&c);print(a"^"b" - "c"^"d) for(n=1,30,print1(n"! = ");nice(lookfor(n!))) ff(t)=for(i=1,#pw-1,for(j=i+1,#pw,if(ispower(pw[j]-pw[i]-t),return([pw[j],pw[i]])))); nice2(n)=my(a,b,c,d,e,f,v=ff(n!));b=ispower(v,,&a);d=ispower(v,,&c);f=ispower(v-v-n!,,&e);if(e>0,print(n"! = "a"^"b" - "e"^"f" - "c"^"d),print(n"! = "a"^"b" + "-e"^"f" - "c"^"d)) April 4th, 2009, 06:52 AM   #3
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Re: Diophantine equation : factorial and powers

Quote:
 Originally Posted by momo Can we express n! as sum of powered numbers? n! = (a^b + c^d + e^f ....) - (a'^b' + c'^d' + e'^f' ....) where a,a',b,b',c,c' ..... are integers > 1 I'm not looking for a trivial solution. The number of the variable used has to be minimal of course. If we can find a general solution then the factorization problem will be over.
Every integer is the sum of four squares. That is nontrivial. Hence yes, you can always write

for some integers a,b,c,d. Or you can use 9 cubes, or 19 fourth powers.

Quote:
 the factorization problem will be over
I am tired of hearing claims like this. You don't seem to be very knowledgeable in mathematics. Usually, when somebody claims to have a solution to an important problem, the solution is wrong. April 4th, 2009, 06:59 AM   #4
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Re: Diophantine equation : factorial and powers

Quote:
Originally Posted by brunojoyal
Quote:
 Originally Posted by momo Can we express n! as sum of powered numbers? n! = (a^b + c^d + e^f ....) - (a'^b' + c'^d' + e'^f' ....) where a,a',b,b',c,c' ..... are integers > 1 I'm not looking for a trivial solution. The number of the variable used has to be minimal of course. If we can find a general solution then the factorization problem will be over.
Every integer is the sum of four squares. That is nontrivial. Hence yes, you can always write

for some integers a,b,c,d. Or you can use 9 cubes, or 19 fourth powers.

Quote:
 the factorization problem will be over
I am tired of hearing claims like this. You don't seem to be very knowledgeable in mathematics. Usually, when somebody claims to have a solution to an important problem, the solution is wrong.
Can you write this FACTORIAL number

16452142516874952231465221452111233522165123415213 65412987411256421236898741554125511122556345413661 123635645198744512456314145514556541!

as sum of 4 squares?

I know we can do in theory but how to solve it practically? April 4th, 2009, 07:04 AM #5 Senior Member   Joined: Nov 2007 Posts: 633 Thanks: 0 Re: Diophantine equation : factorial and powers What I'm talking about is very simple If we can express any factorial of one big big number (100000 digits or more) as : a^b + c^d (for example) then you can factorize any number < to this number. That's it. I'm not claiming nothing. I close this post anyway. April 4th, 2009, 07:13 AM   #6
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Re: Diophantine equation : factorial and powers

Quote:
 Originally Posted by brunojoyal Every integer is the sum of four squares. That is nontrivial. Hence yes, you can always write for some integers a,b,c,d. Or you can use 9 cubes, or 19 fourth powers.
True. I'm reading Nathanson's book on that right now, oddly enough.

But it's actually not quite enough here, since momo requires the base to be greater than 1. 0 isn't a problem, but 1 is. I suppose we can always add/subtract 9-8, though. So I guess you're right after all.  April 4th, 2009, 07:24 AM   #7
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Re: Diophantine equation : factorial and powers

Quote:
 Originally Posted by momo Can you write this FACTORIAL number 16452142516874952231465221452111233522165123415213 65412987411256421236898741554125511122556345413661 123635645198744512456314145514556541! as sum of 4 squares? I know we can do in theory but how to solve it practically?
That number is too large to work with, having
22174514878431004692219364866552514726623549098554 39563429370449573913993660062043775137385374329379 79812945000881840846773887256567649950
digits. But there are efficient methods to decompose numbers into the sum of four squares; see for example
http://www.alpertron.com.ar/4SQUARES.HTM May 8th, 2009, 12:00 PM   #8
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Re: Diophantine equation : factorial and powers

Quote:
 Originally Posted by momo Can we express n! as sum of powered numbers? n! = (a^b + c^d + e^f ....) - (a'^b' + c'^d' + e'^f' ....) where a,a',b,b',c,c' ..... are integers > 1 I'm not looking for a trivial solution. The number of the variable used has to be minimal of course. If we can find a general solution then the factorization problem will be over.
If n>=4, you can write n!=M*N where M and N are even numbers.

Let U = (M+N)/2 and V = (M-N)/2

You get: n!=M*N=(U+V)(U-V)=U^2-V^2

Example for n=7: 5040 = (2*3*4)*(5*6*7) = 24*210 = (117-93)*(117+93) = 117^2-93^2

Best regards,

Dar�o Alpern
Buenos Aires - Argentina May 8th, 2009, 12:21 PM   #9
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Re: Diophantine equation : factorial and powers

Quote:
Originally Posted by alpertron
Quote:
 Originally Posted by momo Can we express n! as sum of powered numbers? n! = (a^b + c^d + e^f ....) - (a'^b' + c'^d' + e'^f' ....) where a,a',b,b',c,c' ..... are integers > 1 I'm not looking for a trivial solution. The number of the variable used has to be minimal of course. If we can find a general solution then the factorization problem will be over.
If n>=4, you can write n!=M*N where M and N are even numbers.

Let U = (M+N)/2 and V = (M-N)/2

You get: n!=M*N=(U+V)(U-V)=U^2-V^2

Example for n=7: 5040 = (2*3*4)*(5*6*7) = 24*210 = (117-93)*(117+93) = 117^2-93^2

Best regards,

Dar�o Alpern
Buenos Aires - Argentina
Thank you for your idea.
Wilson theorem state that if p is prime then (p-1!) mod p = p-1
My idea is :
If we can express (p-1!) as x^2 - y^2 we have to find x^2 or y^2 such as x^2 = z^(p-1) or y=t^(p-1) so we will have to solve only one of the two because z^(p-1) or t^(p-1) mod p = 1 (if p prime)
We have to write a general formula such as we can be sure to isolate z^(p-1) or t^(p-1)
By expanding (p-1)! we can find such form
(p-1)! = z^(p-1) - x^2
or
(p-1)! = y^2 - t^(p-1)

For some factorials it is possible to build such form.

Thank you for your idea. Tags diophantine, equation, factorial, powers factorial diophantine equation

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