114 Mr Pocklington, Some Diophantine Impossibilities. 
say, and hence must be a power of p, say p*, where X + 2k+4+ 1. 
Dividing by < oe using the third lemma we get } 
—ly= + p8, mp? — Any? = + p* ae, ; 
First eae that is even. If we take the lower sign the 
first equation shows that J/=1, for otherwise J would contain a 
prime of the form 4n +3 and could not divide B? + pe which is” 
a sum of two squares the first of which is prime to l. Hence 
=N. The first equation- has become y?=§?+ p*6*, whence y_ 
o odd, and as a and 8 are odd the second equation gives 
N—4=—1, mod. 8, 
which is contrary to the hypothesis. i 
If we take the upper sign the second equation shows that. 
m= 1, for otherwise it would contain a prime of the form 4n+3_ 
and so could not divide 47+ p*a, where 2y is prime to m. 
Hence /= NV and the second equation, 6? = pa? + 4ry?, gives 
B=2 +7’, y= &n, ; 
and on substituting in the first we get (€? + 7)? — NE? = po? 
This is of the original form, for \ being even, p* is a square. Also 
En = < 2y6< y< ay, which contradicts the assumption. ; 
Next suppose that X is odd. Solving the equations for B and 
y we have @ — mé?= + pity’, lo? — 40° = + p*O?, where w= 2h + 1—X- 
and is even. These equations are of the same form as those just 
considered, with J and m, a and £, y and 6 interchanged, and 
in precisely the same way we show that the lower sign gives 
a contradiction of the hypothesis, and the upper sign gives an 
equation of the original form with a smaller value of the product 
of the indeterminates, which contradicts the assumption. Hence 
the equation is impossible under the restriction on WV stated in the 
hypothesis. { 
The most viupeneym case of the proposition is that of V = bi 
or that a+ a*y’+ y* = 2 is impossible. Hence, just as in § 5, an 
integral triangle with an angle of 120° cannot be equal in area © 
to an equilateral triangle. Also 
at + Py? + yt = (a + wy + Y?) (a? — ay + ¥?), ; 
so that both brackets cannot be squares. Hence a parallelogram — 
with integral sides and diagonals cannot have an angle of 60° 
between the sides or between the diagonals. 
If at — 140°? + yi = 2 has any solution it has one in which — 
x is prime to y. If one of these is even we write the equation — 
(22+) — 16ay?=2, whence #+y=w+v, 2ay=w, and <q 
ut + uv? + v! = (a2 — y?)?, which is impossible. If « and y are both — 
odd we write the equation 
2 2 
(- = i — 4aty? = 2/4, 
*) 
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