We treat Generalised Taxicab Numbers problem, Taxicab(5,6,m) again. In the same way as Taxicab(4,3,m), we prove that there are always many solutions for any m. x1.Introduction_{11}^{5}+x_{12}^{5}+x_{13}^{5}+x_{14}^{5}+x_{15}^{5}+x_{16}^{5}= x_{21}^{5}+x_{22}^{5}+x_{23}^{5}+x_{24}^{5}+x_{25}^{5}+x_{26}^{5}= ... = x_{m1}^{5}+x_{m2}^{5}+x_{m3}^{5}+x_{m4}^{5}+x_{m5}^{5}+x_{m6}^{5}Proof. We consider x1^5+x2^5+x3^5+x4^5+x5^5+x6^5 = N. Let x1=t+a, x2=t+b, x3=t+a+b, x4=t-a, x5=t-b, x6=t-a-b. Hence x1^5+x2^5+x3^5+x4^5+x5^5+x6^5 = 20t(a^2+ab+2t^2+b^2)(a^2+ab+b^2)+6t^5. t is arbitrary. Hence let consider the rational solutions of a^2+ab+b^2=n. We know that equation a^2+ab+b^2=n has infinitely many rational solutions for some n. To make the story easy, let n=1. Of course, the same can be said for other than n = 1. Quadratic equation a^2+ab+b^2=1 has a parametric solution below. (a,b)=((k-1)(k+1)/(1+k+k^2), -k(2+k)/(1+k+k^2)). Thus we can obtain distinct rational solutions (a,b) for any m as follows. (a,b)=(n2.TheoremThere are integers that can be expressed as the sums of six fifth powers in any m of ways. x_{11}^{5}+x_{12}^{5}+x_{13}^{5}+x_{14}^{5}+x_{15}^{5}+x_{16}^{5}= x_{21}^{5}+x_{22}^{5}+x_{23}^{5}+x_{24}^{5}+x_{25}^{5}+x_{26}^{5}= ... = x_{m1}^{5}+x_{m2}^{5}+x_{m3}^{5}+x_{m4}^{5}+x_{m5}^{5}+x_{m6}^{5}m is arbitrary._{11}/d_{11}, n_{12}/d_{11}), (n_{21}/d_{21}, n_{22}/d_{21}),...,(n_{m1}/d_{m1}, n_{m2}/d_{m1}). To clear the denominators, multiply the (a,b) by d_{11}d_{21}...d_{m1}. We obtain the integer solution of a_{11}^{2}+ a_{11}b_{11}+ b_{11}^{2}= a_{21}^{2}+ a_{21}b_{21}+ b_{21}^{2}=...= a_{m1}^{2}+ a_{m1}b_{m1}+ b_{m1}^{2}for any m. For large enough number t, we can obtain the positive integer solution. Finally, we obtain the positive integer solution of x_{11}^{5}+x_{12}^{5}+x_{13}^{5}+x_{14}^{5}+x_{15}^{5}+x_{16}^{5}= x_{21}^{5}+x_{22}^{5}+x_{23}^{5}+x_{24}^{5}+x_{25}^{5}+x_{26}^{5}= ... = x_{m1}^{5}+x_{m2}^{5}+x_{m3}^{5}+x_{m4}^{5}+x_{m5}^{5}+x_{m6}^{5}for any m. Q.E.D.@m=5, a^2+ab+b^2=1, t=2 (a,b)=(3/7, -8/7),(8/13, -15/13),(24/31, -35/31),(16/19, -21/19),(33/37, -40/37). (x1,x2,x3,x4,x5,x6)=(17/7, 6/7, 9/7, 11/7, 22/7, 19/7), (34/13, 11/13, 19/13, 18/13, 41/13, 33/13), (86/31, 27/31, 51/31, 38/31, 97/31, 73/31), (54/19, 17/19, 33/19, 22/19, 59/19, 43/19), (107/37, 34/37, 67/37, 41/37, 114/37, 81/37). Finally, we obtain 4816253^5+1699854^5+2549781^5+3116399^5+6232798^5+5382871^5 = 5186734^5+1678061^5+2898469^5+2745918^5+6254591^5+5034183^5 = 5501678^5+1727271^5+3262623^5+2430974^5+6205381^5+4670029^5 = 5636358^5+1774409^5+3444441^5+2296294^5+6158243^5+4488211^5 = 5735093^5+1822366^5+3591133^5+2197559^5+6110286^5+4341519^5.3.Example

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