Mr Pocklington, Some Diophantine Impossibilities. 119 
Therefore we have &n either < y@ or < ya or <88 or <6a, so 
that &)<aByd< xy. But if one of the factors does vanish we 
find on working back that numerically a, B, y, 6 are all equal 
and a = 2y. 
Hence if we are given any solution in which z is prime to y 
except the solution (1, 2) we can deduce a smaller one, and 
repeating the process must ultimately reach the excepted solution 
(1, 2). Hence we can reach any solution by retracing our steps 
from this one. Let then &n satisfy &— 4£%n?+ 1=€?, where 7 
is even and prime to & In retracing the work of the first case 
we see that each number can be determined, and that a is prime 
to 6, then that a, 8, y, 6 are mutually prime in pairs, and finally 
that x is prime to y. In retracing the work of the second case © 
we have 6, c and € mutually prime and c¢ odd. Then 
a/d =(be + €)/(160? — c?). 
If we take a/d to be in its lowest terms we have a prime to b,c 
and d, and d odd and prime to 6 and c. Then (@ + @)/2 is prime 
to (@ —a)/2 and a is prime to @, both being odd. The previous 
equations now show that a, 8, y, 6 are all odd and mutually 
prime, so that z is prime to y. The ambiguous sign in the 
value of a/d may be taken either way unless bD=c=C=1 
numerically, and the values of a/d are not reciprocals of each 
other. Hence in general we can deduce three solutions from 
any given one, but only two can be deduced from the solution 
elt 2). 
All these solutions are different. For if in retracing our steps 
we arrive at the same solution in different ways, then in the 
direct work there would be two alternative ways of proceeding. 
But this does not happen, for the case that the work falls in can 
be determined by the remainder of a+ 7? to divisor 8, and in each 
ease each step is uniquely determined. 
By a method similar in principle but differing in detail we 
can find the complete solution of 2a*—y*=z. From the least 
solution (1, 1) we can deduce one other, but from any other we 
can deduce two new ones, and by continuing the process can 
reach any solution. In the case of y!—2a*= 2? we deduce one 
solution from each solution of the allied equation 2a4 — y*= 2 and 
from any solution of the equation itself we can deduce one other, 
and continuing can reach any solution. The arrangement of the 
primitive solutions is not a dichotomously or trichotomously 
branched one, as before, but consists of an infinite number of 
unbranched sequences. 
13. If the equation 2” +4” = 2 has any solution it has one 
in which # is prime to y, one of them, say y, being even. Then 
a =u—v, y=Quv, and if u is even u+tv=a", u—v=", 
