116 Mr Pocklington, Some Diophantine Impossibilities. 
of each other is satisfied. It is sometimes convenient to com- : 
mence by writing the equation in the form (a?— yf? + Nay=2, — 
and then we must examine the equation to see if e=y gives a 
solution. Sometimes we have to use the method developed in § 12, — 
second case. Collecting results, we have a+ na°y?+y*=2 im-_ 
possible if n is 0, 1, 3, 4, 5, 6, 7 (unless =y), 9, 10, 11, 14 (unless — 
z=), 15, 18, 19, 20, 21, 22, 25, 28, 29, 35, 45; 51, 59, 65, G9, 74am 
81, 91, and if —n is 1 (unless x = y), 3, 5, 6, 7, 8, 10, 12, 14, 17, © 
18, 19, 20, 21, 22, 23, 24, 27, 29, 31, 45, 54, 55, 60,61, 69) (am 
If n lies between — 30 and 30 the equation can be solved except — 
in the cases just given. If n= —15 the least solution is = 95, — 
y = 24, but in the other cases the values of # and y in the least 
solution do not exceed 6. 
10. We now propose to prove the impossibility of 
(a? — 5y?)? + 128277? = 2°, 
not so much for its own sake as because of a deduction we make 
from it. As before, if it has a solution, take that in which ay 
has the least value. Then a is prime to y. If # is divisible 
by 5 let it be 5z’, then the equation becomes | 
(y? — 5a”)? + 128 y2a"? = (2/5)?, 
which is of the same form as the original equation but with a 
smaller value of the product of the indeterminates. Hence w is — 
prime to 5 and a? — 57? to xy”. | 
Firstly, suppose that 21s even and y odd. Then a?—5y?=2v?—v?, 
Axy = uv, where wu is odd and v even, x=a8, y=y¥6d, u=ay, 
v=486 and a? (6?4+ 9°) = &(328?+4 57”), where 8 is even and y 
odd. The only possible greatest common divisor of the brackets 
is unity, so that a?=32@?+ 5y?, which is impossible to modu- 
lus 8. 
Secondly, suppose that zis odd, and y even. Then - 
e— by =w—20, 42y= U0, 
where wu is odd and v even, 7=a8, y=y6, u=ay, v=486, where 
a, 8, y are odd and 6 even and §?(e?+ 320°) =9?(a?+ 56’). The 
greatest common divisor of the brackets divides 27, hence 
a? + 3262= 9? or = Or’, a + 56?= 8? or = 96", the other cases being 
excluded by using the modulus 8. Since 6 is even the second — 
of these equations gives 1n either case + a= &?— 57, 6 = 2&n, where 
— or 7 is even, and the first becomes (£— 5n?)? + 128&n? = sq. 
This is of the original form, and 2) =6 $y < 2ay, which is 
impossible since we have chosen a solution for which wy has the 
least possible value. Hence the equation has no solution in — 
which w and y are one even and the other odd, or in which both © 
are even but containing 2 to different powers. 
