a theorem of Dirichlet 113 



where V is one particular unity of the order, and m is a positive 

 or negative integer or zero. 



In what follows we shall restrict ourselves to the positive sign 

 on the right-hand side of equation (i), since the negative sign 

 merely replaces {x, y, z) by (— x, — y, — z). Also when x, y, z are 

 all positive, we shall refer to {x + 2/'^ -f s^-) as a " unity of positive 

 integers ". 



Suppose that 



r = *• + 2/^ + z"^" 

 is any unity of the order of '^ : we shall first prove the following 

 inequalities, viz. : 



|a--2/^|, JT/^-^^-j, |^^-^-a;!^2/V(3r) (ii). 



For, if we write a-^- x — y*^, 



^ = y^-z^\ 



and 7 = z^- — X, 



we see that the equation satisfied by x, y, z can be thrown into 

 the form 



r(a^-h/3-^ + 7^)=2: 



so that we have 



a-^+/3^ + 7' = 2/ri 



and a + /3 + 7 = ]' 



From these two equations, assuming F to be constant, we find 

 that the maxima and minima for each of a, /8, 7 are 



± 2/V(3r) ; 



from which the truth of the statement (ii) follows immediately. 



Further, we have the fact that if F = a; + 2/^ + z^" is any unity 

 of the order and F> 1, then x, y, z will be positive. 



For, since F> 1, we have the inequalities 



\x-y'^\,\y'^-z'h^\,\z'^^-x\< 2/V3 < US. 



But, r being positive, the only possibilities of negative signs 

 occurring amongst x, y, z are either (a) one negative and two 

 positive or (6) two negative and one positive ; and in each case 

 two of the inequalities given would take the form 



I X, + yU<^ I < 115, 



where \ and fi are positive integers and <^ ^ \/2, which is obviously 

 impossible, except in the trivial case of one or more of the quantities 

 x, y, z vanishing : it will be seen, on examining the inequalities, 

 that the only possibility is x = l, y=0, z = 0, which gives r = l 

 and so is excluded. Hence x, y, z must be positive. From this 



