368 report— 1865. 



If we write 2m' + d'=x, o'—2m'=y, 2m'+d'—d'e=z, so that conversely 

 2m'=x— ?/— z, d'=y+z, B'—x—z, the equation m=2m' 2 -\-d'B' becomes 

 2m=x 2 +y 2 — z 2 , the indetermiuatcs being subject to the conditions 



y + z>0, z<x. 



If in addition r-f- x <0, the conditions are satisfied by the two solutions 

 [x, y, as], \_—ai s y, as]; which give rise, in the sum ~Zf(d" + 2m'), to two terms 

 which cancel one another. We need only therefore consider those solutions, 

 which satisfy the inequalities, ;/+r>0, .r + r>0, z<x, or, which is the same 

 thing, if [z] represent the absolute value of z, 



y+z>0, .r>0, [*]<* (d) 



Again, if we write ^(d l +B l )=as, '»,=?/> i( ( h~^i) =z > tne equation 

 2m=m 2 + d l S 1 becomes 2m=x 2 + y 2 — z 2 , the iudeterminates being subject to 

 the conditions 



y>0, .r>0, |>]<.v (d') 



To estabhsh the proposed equation it is now only necessary to show that 

 the equation 2m=x 2 + y 2 — z 2 admits of equal numbers of solutions satisfying 

 the inequalities (d) and (d'). But this is evident ; for if [x, y, z] satisfy one 

 of the two sets of inequalities, but not both, [x, —y, —z~] satisfies the other, 

 but not both. 



By combining these two lemmas it may be proved that four times the number 

 of solutions of the equation 



2 a+l m=m{ + d 2 Z 2 + (d 2 + L)B 3 (A) 



(in which m is a given uneven number, and a a given exponent > 0) is 

 A(2 a+l 7ii) — r(2 a+5 m)— r'(2 a+I m), the indeterminates m lt d 2 , B 2 , B 3 being all 

 positive and uneven and d 2 —o 2 being evenly even. Representing by m' any 

 number whatever, and by d' , h', d 3 positive uneven numbers, let us consider 

 the two equations 



2 a+l m-2d 3 ? s =mi + d 2 Z 2 , (e) 



2 a m-d 3 B 3 =2m' 2 + d'l', (e') 



and let/(#) be an even function, so that f(x —d^) —f(x + d 3 ) is an uneven 

 one, and may be used instead of /(.*') in the second lemma. Applying that 

 lemma to the two equations (e) and (e), and afterwards summing for eveiy 

 value of d 3 , we find 



{/(^-".)-/(^+".)] 



= S[/(2» l ' + d' -d 3 ) -f(2m' + d' + d 3 )l 



the summations extending to all solutions of (e) and (e) respectively. Ob- 

 serving that if m' = 0, /(2>»' + .r) is an even function of x, and that if ml is 

 not =0, f(—2m'+x)+f(2m'+x) is an even function of an, we transform 

 the second member by the first lemma, and we obtain 



s[,(^-*,)-,(i±iM)] 



=2v-'v f Z[/(2m')-/(2^+2m')], 

 the second summation extending to every solution of the equation 



2 a m-2m'-=2'<d%, 

 d and <S being positive and uneven, and 2 V representing the highest power of 



