and Cubic Binomial Congruences with Prime Moduli 59 



5. Let 8 be the arithmetical cube root of a, which we 

 assume* not to be a cube. Findf ti, Ui, Vi such that the norm 

 N = t{^ + au^^ + a'-Vi" — SatiU^i\ of the algebraic number 



is a cubic non-residue of p. We see that as a is a cubic residue 

 of p we have U^' = ti-\- UiS+ i\8'^, so that if 



U^'-^ = tp-i + Up-i 8 + %_i 8- 



we have iip^i = Vp_i = 0. Now taking ?7'" where m is in turn 

 (p— l)/3, (p — l)/9, etc. we see that we cannot always have 

 u,n = v„i = 0. Let f/""^ be the last of this series for which this 

 happens. Then m is divisible by 3, for otherwise the norm of [7"\ 

 which reduces to t^, would be congruent to the non-residue iY'". 

 Putting m = on we have 



tsn = tn + <Kln + a-^n + QatnUnVn, 

 = Usn = 3 (tn'Un + ^^tnVn' + «",rWn), 

 = V.,n = 3 {taUn' + tuVn + aUnVn'')- 



The last two give tn{av^i"' — Vn)= 0; so that if tn is not divisible 

 by p we have w = Un/v,,. as one solution of n^ = a, for as Un and v,i 

 are not both divisible by p this shows that neither is. They also 

 give Vn (ait^n — tn) = 0, and so w = tnjihi is a solution. Eliminating 

 a from the same two congruences we see that the ratio A, of the 

 two xs, satisfies V + X + 1 = 0, so that they are distinct. The 

 third solution follows immediately. 



If however tn is divisible by p the two congruences show that 

 either Un or Vn is divisible by p. We now have rtM,i,^=iA'" or 

 a-Vn = -Y". In either case n must be divisible by 3 as before, and 

 we have as one solution x = N^'jun or x = aVnjN'' respectively, 

 where r = n/S. 



* Simply because of the way in wliich for the sake of shortness we are stating 

 the method. 



t This again must be done by trial. In order to use the Law of Cubic 

 Beciprocity we must express p in the form ii' + xiv + v'^, which requires the solution 

 of ^2 + 3 = 0. 



5—2 



