[An easy and rapid program for solving a^3+b^3+c^3+d^3=0 in UBASIC.]
1000 input A : B=1 : N=3
2000 K=A
2400 S=-K^N+A^N+B^N
2440 if 0<=S then S1=S-int(S^(1/N))^N else S1=S+int((-S)^(1/N))^N
2441 if and{S1=0 , S<0} then print A ; B ; (-S)^(1/N)
2442 if and{S1=0 , 0<S} then print A ; B ; "-" ; S^(1/N)
2700 if 0<S then A=A-1 else B=B+1
2720 if A<B then end
4000 goto 2400
How to use :
run
input 12^6
output
2985984 1 - 1
2985912 124415 1
2985408 248816 16
2981376 497408 256
2952868 956832 14684
2949120 991232 4096
2933496 1113183 6561
2890152 1353935 14641
2827800 1588847 28561
2518400 2200192 - 6144
It means :
12^18 = 2985984^3 + 1 - 1
= 2985912^3 + 124415^3 + 1
= 2985408^3 + 248816^3 + 16^3
...
...
Noam Elkies said that this program runs as fast as his p-adic method.
[A sequence of identities for a diophantine equation : A^3+B^3+C^3=D^3 .]
The iteration for the identity is as follows :
x0=9*a^4; y0=-9*a^4+3*a*b^3; z0=-9*a^3*b+b^4; u0=b^4
x1=9*a^4; y1=-9*a^4-3*a*b^3; z1=9*a^3*b+b^4; u1=b^4
for n=2
xn=(432*a^6-2*b^6) * x_(n-1) - x_(n-2)*b^6 - 108 * a^4*b^6
yn=(432*a^6-2*b^6) * y_(n-1) - y_(n-2) *b^6- 108 * a^4*b^6
zn=(432*a^6-2*b^6) * z_(n-1) - z_(n-2)*b^6 + 216 * a^6*b^4 + 4*b^10
un= b^(6*n-2)
for 3<=n
xn=(432*a^6-2*b^6) * x_(n-1) - x_(n-2)*b^12 - 108 * a^4*b^(6*n-6)
yn=(432*a^6-2*b^6) * y_(n-1) - y_(n-2) *b^12- 108 * a^4*b^(6*n-6)
zn=(432*a^6-2*b^6) * z_(n-1) - z_(n-2)*b^12 + (216 * a^6*b^4
+4*b^10)*b(6*n-12)
un= b^(6*n-2)
for all n
xn^3+yn^3+zn^3=un^3
ex.
If n=4, the identity is as follows :
(725594112*a^22-31912704*b^6*a^16+194400*b^12*a^10-279*b^18*a^4)^3+
( -725594112*a^22-241864704*b^3*a^19-8398080*b^6*a^16+2799360*b^9*a
^13+85536*b^12*a^10-7776*b^15*a^7-153*b^18*a^4+3*b^21*a)^3+
( 725594112*b*a^21+120932352*b^4*a^18-8398080*b^7*a^15-839808*b^10*
a^12+23328*b^13*a^9+1296*b^16*a^6-9*b^19*a^3+b^22 )^3=b^66
If b=1, then it becomes the same thing as the iteration on
"Identities Of Equal Sums Of Like Power" :
[A solution of Fermat's equation ?]
5949503719112230284735150781047420975571215969633783660978692570826281101309189803771999166533150006397466555165181072458877^3
+693167423530203568796466556841646172413753795462810020862370805265266981207049609344143725917800499174450944129560853125400625152^3
-12957693805689906522723235353228014067942008670140322824814667098806805313380604081166444397018867983934899325^3
=2^1284
2^1284 is 2-adically small, so this is almost an example of a^3+b^3-c^3=0
[ Generalized Taxicab Number ]
T(k,m,n) :
the smallest number such that representable as a sum of m k powers in n ways.
ex.
T(3,2,n) is the same thing as Ta(n) which is known as Taxi cab Number :
the smallest number representable as a sum of two cubes in n ways.
for n<=5 , they are known.
[The smallest records for T(3,m,n)]
T(3,2,6)=Ta(6)<=8230545258248091551205888
= 11239317^3 + 201891435^3
= 17781264^3 + 201857064^3
= 63273192^3 + 199810080^3
= 85970916^3 + 196567548^3
= 125436328^3 + 184269296^3
= 159363450^3 + 161127942^3
. ( David W. Wilson )
I have two examples.
T(3,3,16)<=810000^3
= 809730^3+80991^3+9^3
= 807840^3+161856^3+144^3
= 802710^3+242271^3+729^3
= 792720^3+321696^3+2304^3
= 776250^3+399375^3+5625^3
= 766200^3+433800^3+6000^3
= 751680^3+474336^3+11664^3
= 656400^3+628800^3+14400^3
= 796500^3+295500^3+69000^3
= 795000^3+306000^3+69000^3
= 720000^3+540000^3+90000^3
= 782550^3+373725^3+27675^3
= 788355^3+346140^3+15165^3
= 783000^3+371250^3+60750^3
= 724140^3+532980^3+68040^3
= 736306^3+509492^3+6366^3
T(3,4,15)<=1350000^3
= 1349538^3+136140^3+12360^3+462^3
= 1346888^3+256987^3+15397^3+728^3
= 1346737^3+261085^3+7421^3+221^3
= 1339010^3+390610^3+13034^3+466^3
= 1333215^3+449190^3+12330^3+225^3
= 1285830^3+694124-3+12772^3+762^3
= 1349200^3+163440^3+17640^3+1880^3
= 1346760^3+260280^3+34020^3+540^3
= 1342800^3+339570^3+11070^3+540^3
= 1339160^3+388750^3+34930^3+1880^3
= 1337000^3+412880^3+22660^3+1180^3
= 1330500^3+471870^3+25440^3+1170^3
= 1311240^3+590490^3+14310^3+960^3
= 1235150^3+832050^3+11400^3+1000^3
= 1230230^3+842700^3+29680^3+2010^3
[A Fermat like equation]
Hello, number theorists.
I have considered about the following equation :
[c^r] = [a^r] + [b^r] , c,a,b, are integers and r is real number. [x] means integer part of x.
if r is integer, then it becomes Fermat's equation.
At first, I conjectured that the number of solutions behaves as follows
:
a point k exists and,
r<k .... solutions exist
k<r .... no solution exists.
k should be between 2 and 3. Because c^2=a^2+b^2 has solutions and c^3=a^3+b^3 has no solution.
But it seems to be more complicated.
I found these solution :
r=2.1
c,a,b = 14353, 14351, 298
10101, 7904, 6549
r=2.15
c,a,b = 10101, 8589, 5719
r=2.2
c,a,b = 3^10, 58883, 5847
r=2.7
c,a,b = 50, 47, 25
53, 41, 41
1060, 1059, 116
r=3.0001
c,a,b = 89, 86,41
r=3.001
c,a,b = 389, 383, 139
r=3.4
c,a,b = 317, 313, 125
Is there any study for this problem?
I appriciate you any information about it.
Yasutoshi