108 Mr Pocklington, Some Diophantine Impossibilities. 
Some Diophantine Impossibilities. By H. C. PockLineToN, | 
M.A., St John’s College. 
[Recewed 28 October 1912.] 
1. THE object of this paper is mainly to discuss some 
equations obtained by equating a quadratic function of # and 7 
to a square, either to show that they are impossible in integers or 
rational fractions (§ 3—§ 10), or to completely find their solutions — 
in integers (§ 12). Two theorems on arithmetical progressions of 
which given terms are to be squares are given in § 6, 11, and the 
impossibility of a” + y” = 2 is discussed in § 13. 
2. We make use principally of three lemmas. The first is 
that if zy= 2", and w is prime to y, then both w# and y are nth 
powers. The proof is obvious on expressing z as a product of 
primes and their powers. For each prime that occurs in z must 
occur in # or y, but not in both, and clearly occurs raised to a 
power the index of which is some multiple of n. If n=2 we 
have an alternative proof in Euclid 1x. 2, and if n is a power of 
2 we can get our result by a repeated application of Euclid’s 
proposition. We have assumed that w and y are positive. If 
they can be negative there is the alternative =— wu y=—v". 
By a repeated application we see that if zyz= w”, then a, y and z 
are nth powers. 
We shall also require the following application of the lemma. 
Let a? + Ny? = 2, where either x or zis prime to Vy. Then w# is 
prime to z, for if p is a prime common divisor of # and 2 it 
divides Vy? and hence Vy. Hence the greatest common divisor 
of z+ and zg—« is 1 or 2. First suppose that it is 1, ie. that 
Ny is odd. Then, writing the given equation Ny? =(z+)(z—2), 
we see that z+a=lu, z—x=mwv, where lm=WN and lw is prime 
to mv. We now have y?= w whence u = &, v= 7? and so 
a = (IE? — map)/2, y= En, 2 = (E+ myp)/2. 
Next suppose that the greatest common divisor is 2, 1e. that V7 
is divisible by 4. Then if y is even we have 
N (y/2) = 4 (2 +2).3 (2-2), 
where (z+ «)/2 is prime to (¢—«)/2. Treating this as before we 
find # = 1E*—m7n?, y= 2&n, 2 =1E?+ mn. If y is odd, N must 
be divisible by 8, and N7?/4=4(¢+a).4(z2—2), whence as 
before we find «=/&?— mm’, y=&, z=l1E?-+ my’, where 1é is 
prime to my and Im= N/4. 
