[ Mr Pocklington, Some Diophantine Impossibilities. 117 
Lastly, let both # and y be odd. Then 2 — 57? is divisible 
by 4, the quotient being odd. Writing the equation 
2 Bap\2 
(* 7) + 8474? = sq., 
we have (a? — 5y)/4 = + (w? — 2v"), zy = wv, where wu and v are odd, 
and =a, y=70, w=ay, v=, where a, 8, y, 6 are odd. 
‘Taking the lower sign in the first equation we have 
a (02+ Ary?) = & (SB? + 5y?”), 
The greatest common divisor of the brackets must be unity, so 
that a?=86?+ 5y?, which is impossible to modulus 8. Taking 
‘now the upper sign we have £?(a?+ 86?) = y?(40?+ 58"). The 
greatest common divisor of the brackets must divide 27, hence 
4+ 88°= 9? or =9y?, 40?+50?= 8? or 96", the other cases being 
excluded by using the modulus 8. The second of these gives 
+a=(E?—5n’)/2, 5=&n, where & and n are odd, and the other 
becomes 
2_ Fk ne 
(Ft) +860 =, 
which is of the original form. Also &n = 6 = ay/aBy < vy unless 
a=S=y=1 numerically. But then e=1, y=d, w=1, v=6 
and (a —5y*)/4= 1? — 20 gives 6&?=1, so that our solution is 
-£2=y=1 numerically. Hence if we have any solution in which 
z and y are odd and not both = +1, and «# is prime to 5y, we can 
find another satisfying the same condition, in which the product 
of the indeterminates has a smaller value. This is impossible. 
Hence the given equation can be satisfied only if e=y or «=5y 
“numerically, or if # or y vanishes. 
) 11. We deduce that the first, second, fifth and tenth terms 
of an arithmetical progression cannot all be squares, unless the 
first term is zero or all are equal. For if possible let the squares 
be those of 6, y, 8, a(supposed positive). Then 
a? B? au 5ry?o? a AaPry? ae 820” 
and if r=a8, y=¥9, u=ay, v = BO, we have a? — 5y?=4 (w — 20°), 
ay=uw and (a? — dy)? + 12847? = 16 (u?+ 2v’)?. This is impos- 
sible except in four cases. If «=y we have a8=y6, which gives 
— 4928? = doy? — 828. Eliminating y this gives 06? = 26 — a’, and 
so a?= 6, whence the squares are all equal. If e=5y we have 
a= 5y6, which gives 5y%6?=o'y?— 26° Eliminating a this 
gives 5By?= 25y*— 24, which has no solution in integers. It 
a=( or y=0, one of the numbers a, 8, y, 6 must vanish, and 
we easily see that it must be 6. Hence we have the theorem 
stated above. 
