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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[ik]==t,return([pw[i],pw[ik]])));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[1],,&a);d=ispower(v[2],,&c);print(a"^"b"  "c"^"d) for(n=1,30,print1(n"! = ");nice(lookfor(n!))) ff(t)=for(i=1,#pw1,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[1],,&a);d=ispower(v[2],,&c);f=ispower(v[1]v[2]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  
Senior Member Joined: Nov 2007 Posts: 258 Thanks: 0  Re: Diophantine equation : factorial and powers Quote:
for some integers a,b,c,d. Or you can use 9 cubes, or 19 fourth powers. Quote:
 
April 4th, 2009, 06:59 AM  #4  
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Re: Diophantine equation : factorial and powers Quote:
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  
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 Quote:
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 98, though. So I guess you're right after all.  
April 4th, 2009, 07:24 AM  #7  
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 Quote:
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  
Newbie Joined: May 2009 Posts: 2 Thanks: 0  Re: Diophantine equation : factorial and powers Quote:
Let U = (M+N)/2 and V = (MN)/2 You get: n!=M*N=(U+V)(UV)=U^2V^2 Example for n=7: 5040 = (2*3*4)*(5*6*7) = 24*210 = (11793)*(117+93) = 117^293^2 Best regards, Darío Alpern Buenos Aires  Argentina  
May 8th, 2009, 12:21 PM  #9  
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Re: Diophantine equation : factorial and powers Quote:
Wilson theorem state that if p is prime then (p1!) mod p = p1 My idea is : If we can express (p1!) as x^2  y^2 we have to find x^2 or y^2 such as x^2 = z^(p1) or y=t^(p1) so we will have to solve only one of the two because z^(p1) or t^(p1) mod p = 1 (if p prime) We have to write a general formula such as we can be sure to isolate z^(p1) or t^(p1) By expanding (p1)! we can find such form (p1)! = z^(p1)  x^2 or (p1)! = y^2  t^(p1) For some factorials it is possible to build such form. Thank you for your idea.  

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