112) = Mr Pocklington, Some Diophantine Impossibilities. 
a+ 6 and a+ 40? is some divisor of the determinant formed 
from their coefficients, which is 3. But 3 does not divide o?+ 8 
hence the brackets are prime to each other, and applying the 
third lemma we have 02+ &=9?, a? +45°= 6. The last gives 
a=&—7, 6=&n which aris ced in the previous one gives. 
&*— £7? 4+ n+=9%. This is of the original form, for & or 7 is 
even. Also £7 =6 <2y6< y< ay, which contradicts the assump 
tion, and we conclude that the equation is impossible if a or y 
is even. | 
Secondly if « and y are both odd, we again write the equation | 
(2 — w+ ay? = 2? and unless 2—y=0 we find 2—y?=2uy,, 
cy =u?—v*, whence s 
ut — yy? + yt = C= ve + uy? = (x a yy, 
which is of the form just proved to be impossible, for w is prime » 
to v, one of them is even, and (a+ y?)/2 is integral. Heneg 
w — oy? + yt = 2? is impossible uMleSs ic — 
From this we can prove that an integral triangle with an 
angle of 60° cannot be equal in area to an equilateral triangle, 
unless it is itself equilateral. For this requires ¥ 
eP—xsy+y= 2, cy=w*, 
where «, y are the sides about the angle of 60°. If these 
equations have a solution they have one in which « is prime to y, 
and then the second equation shows that s=w?, y=v?, and the 
first becomes w!— uv? + 0'=2?, which is impossible unless w=, 
that is, e=y. ; 
6. If we attempt the second part of the discussion of 
at — ay? alt yf! — 72 
by the same method as was used for the first part we get 
207 1O —= ay a)207— Bo, 
whence @, y”, 8, 6 are in arithmetical progression. It is possible 
but troublesome to go on and complete the proof. As however 
we have proved the theorem in question we can invert the 
present proof and show that four squares cannot be in arith-— 
metical prog wecelen. For if 2, y?, 2, w? are such squares, we have 
2a°2? — 2yw? = ay? — zw, Hence putting | 
U=22, v= yw, E=(ay+zw)/2, n=(ay —zw)/2, 
which we easily see to be integral, we have 
W—v=2En, uy = E?— 77, 
and hence ut —wv?+ vu! = (+77). Also unless the squares are 
all equal (when they can hardly be a progression) we have 
y>", w>zZz or y<a, w< 2, so that wu and v are unequal, and the 
