58 Mr Pocklington, The Direct Solution of the Quadratic 



3. Put D=—a, so that we have to solve cc^+ D = where D 

 is positive or negative but not divisible by p. Let t^ and Mj be so 

 chosen* that t^ — Du-^- = N is a quadratic non-residue of p, and 

 let 



tn = [{t, + U, ^Dr + (t, - U, sjDY]l% 



These numbers are clearly integral. Also 



by use of* which (at first with m = n) we can find the remainders 

 of tn and Un to our modulus without serious difficulty even when 

 n is large. We also have tn — Dun = N'^. 



Supposing that p is of the form 4m + 1, we have D a quadratic 

 residue of 'p, and tp = t^P = ^i, Up = u-pD^P~'^''i- = u^ ; and now 



ti = tp^^t-^ + Dup_^Ui^, tij = tp_iUi + t-^Up_^ 



give on solution tp_^ = 1, Up_-^ = 0. Let p-l = 2r. Then 



= l/^_i = 2trUr 



shows that either t,. or u,. is divisible by p. If it is Ur we put 

 r = 2s and proceed similarly. We cannot have every u divisible 

 by p, for u^ is not. We cannot be stopped by having u,n = with 

 m odd, for we always have 4,,^ - Dum'' = N''\ and this would then 

 give ^,„' congruent to a non-residue. But if m is even we can 

 .proceed further. Hence when we are stopped we must have 

 t,n = 0. This gives - Bii,,^ = N''\ and as - D is a residue m must 

 be even. Putting m = 2n we have = ^^ = tn' + Bun', so that the 

 solution of a^ + D = is got by solving the linear congruence 



UnX =z + tn- 



In applying the method, if n is the largest odd number con- 

 tained m p-l we first work to get the suffixes n, and then the 

 suffixes 2n, 4<n, 8n, etc. Thus in the case of cc'' + 2 = 0, mod. 41, 

 we see that ^j = 3, Ut, = 1 is suitable, and we find t. = 11, u., = 6 ; 

 t, = 29, u,= 9; t,= 23, u,= 15; t^o = 36, u,o = 34 ;' t.^ = 0. ' The 

 solution of 34^ = 36 is a; =30; and so the two solutions of 

 *•- + 2 = are .x=± 30, mod. 41. 



^4. If p is of the form Sm + 2 the only solution of x' = a is 

 a; = l/a'«. If p is of the form 9m + 4> one solution is cc = !/«'"■ and 

 if of the form 9m + 7 one is .x = a^+\ The other solutions are 

 got frojn this by multiplying by (- 1 + 6)12 and (- 1 - e)/2, where 

 t/- + 3 = 0, a congruence which we have shown how to solve. 



* We have to do this by trial, using the Law of Quadratic Eeciprocity, which 

 " •! ?f i'" *^J® method. But as for each vakie of n half tire vahies of / are 

 .suitable, there should be no ditlficulty in finding one. 



