Mr Pocklington, Some Diophantine Impossibilities. 111 
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P=w+ pr, +y=w—pr, where w is prime to py and one is 
even. But by taking residues to modulus 4 we see from the first 
equation that v is even, and that the upper sign must be taken in 
the second. Hence solving, 2pv?= a?— y?, 2w’=a?+y*. Writing 
| 2 = 2 
the second equation (* 5 2 +f ( 9 2 =u?, where «—y>0, we 
Ihave (# + y)/2 = &?— nf, (a ¥ y)/2 = 2&n, whence 
| p (v/2)= En (E? — 77). 
Now & is prime to 7 and so each is prime to €*—7. Hence 
fa po, n= 2, &—a = or E=o?, n= ph, — 7 =o or E=a?, 
n= 8, &—7?=py according to the factor that is divisible by p, 
nd in each case a, 8, y are prime to each other. These give 
Pot — Bi=y, at—pBt=o? or at—Bi=py? respectively. The 
first of these is impossible for p cannot divide B*+ +. The third 
is of the form of the hypothesis, and a®G? > a?B?y? < vw < a < ay? 
which contradicts the assumption that zy had the least possible 
value. In the second « is prime to p, for if not we can write the 
equation p’a*— 6'=y? so that @ is divisible by p, which is 
impossible as a is prime to 8. Hence as a or 8 is even, 
| aty=M, a? Fy = ppt, 
and adding we have A‘ + p’u4 = 2a”, which is impossible, for p does 
not divide a and 2 is a non-residue of p. Hence the given 
equation is impossible. We have proved one case of the theorem 
that « — p®y!= 2 is impossible, and the other case gives 
e=uet+ v, py? = U2 — v2, 
‘so that u‘—v! = pay’, which is impossible. Hence #— p’y*= 2? is 
impossible. 
We have also proved that ay (a — y”) = pz? is impossible if # 
is prime to y, and we easily see that this restriction can be 
removed, hence the area of an integral right-angled triangle 
eannot be the product of a square and a prime of the form 
8m + 3. 
+45, Tf at— ay? + y= 2 has any solution it has one in which « 
is prime to y (and «+y unless e=y=1). Firstly let # or y be 
even, say y, and suppose that we have that solution in which xy 
has the least value, subject to y even. Writing the equation in 
the form (a? — y?)?+ a?y?= 2 we see that a®— y° is prime to ay, so 
that a2— = u?—v, xy=2uv, where the first equation shows 
that v is even. Dividing the last equation by 2 and using the 
second lemma we have w=a8, y=2r76, u=4ay, v= 86, where 
a, 8, y, 8 are prime to each other in pairs, for x is prime to y, and 
u tov. Also a, 8, y are odd and 6 even. Substituting in the 
previous equation we have A? (a+ &)=7 (a + 46°). Now £ is 
prime to y and as @ is prime to 6 the greatest common divisor of 
