Mr Pocklington, Some Diophantine Impossibilities. 118 
last equation is impossible. Hence four squares cannot be in 
Also «4 — 4ay + 2a*y?+ 4ayity', x14 2a? + y, 
x + 4ay + 2a°y? — 4aey?+y* and x*+ 8a°y + 2a°y? — 8ey* + y! 
are in arithmetical progression, and the first three are squares. 
Hence we conclude that the last expression cannot represent a 
‘square ft. 
If a+ 1407? + yt=2 has any solution it has one in which x 
is prime to y. If the indeterminates are odd and even we have 
(2-7) + 162°y?= 27, whence #°—7?=u?—v*, 2ay=uv and so 
Ww — we + vt = (22+ y"), which is impossible, for w and v are odd 
and even and hence unequal. If # and y are odd and unequal, 
Ala 
: ; Tt — UENS z\? 
we write the equation ( Z } + ey? = (j) , Whence 
(2 — y?)/4 = 2uv, vy =v? — v’, 
0 that ut+ 14wv? + ot = (a+ y’)/4, where w is prime to v and one 
of them is even, which we have just shown to be impossible. 
Hence a+ 14a°y?+ y= 2 is impossible unless # = y. 
7. If possible let (a+ y?? — Na*y? = 2, where NV is odd, not 
of the form 8n+3 and not divisible by any prime of the form 
4m + 1, and moreover NV — 4 is an odd power of a prime? (it cannot 
be an even power for NV cannot be the sum of two squares), and 
let us suppose that we have that solution in which ay has the 
least value. Then z is prime to y, and by taking the remainders 
to modulus 8 we see that # and yare not both odd. Let y be 
the even one. Also from its composition V is prime to a?+y’, 
hence the terms of the equation are prime to each other, and 
from the first lemma 2+ y?= lu?+ mv, vy =2uv, where lm = N 
and / is prime to m. Applying the second lemma to the second 
equation we get z=a8, y=2yd, u=ay, v= 6, where a and PB 
are odd, y or 6 is even. Since # is prime to y we have @ prime 
to y, 6 and @ also prime to y, 6, and since Ju is prime to mv, we 
have a prime to 8 and y to 6, and also J prime to 8, 6 and m to 
a, y. Substituting in the previous equation we get 
| a (B® — ly?) = 8 (mp? — dy). 
| 
Now a is prime to 4, and since @ is prime to y, the greatest 
common divisor of the two brackets divides 
| 1, -—l 
i ‘ ’ = N—4= y*t1 
| m, — 4 : 
* Fermat, ‘Inventum novum,” p. 20. 
+ Ibid. p. 36 
4 at Ni then N-4=-~3, This case is included in the proof, taking p=3 
ater. = eens 
; =s 
VOL. XVII. PT. I, 8 
