Mr Pocklington, Some Diophantine Impossibilities. 109 
In the case of a?+ y?=2 we see by taking remainders to 
modulus 4 that w or y is even. Supposing that y is even the 
solution can be put into either of the equivalent forms #=w* — 2, 
Y=2uv, z=wt+ey or w=w, y=(W—v)/2, 2=(wt+v)/2. 
We have assumed that 2 is positive. If it can be negative, 
we must prefix the ambiguous sign to its value. The proof given 
above shows that it is necessary that x, y, z should be of the 
forms found; we easily see by Algebra that it is also sufficient. 
Our second lemma is that if wy=uv then x=a8, y=96, 
w= ary, v= 5. This also may be proved by expressing w and y 
as products of primes and their powers, or more readily by 
noticing that a solution is to take a to be the greatest common 
divisor of # and wu, and put B=a/a, y= u/a, which substituted in 
ay=w gives By= qv with 8 prime to y. Hence A divides v. 
Take § = /8, then this equation gives y= 6, and the lemma 1s 
proved, for we have already arranged that w=a8, u=ay, v= BS. 
Tt is clear that the lemma still holds if a, y, u, v may be negative. 
The extension to the product of m indeterminates equated to the 
product of n others is obvious. 
For an example take the equation a?+y?=u?+v. By finding 
the remainder to modulus 4 we see that # and y are both even, 
both odd, or even and odd, according as u and v are. Hence we 
may suppose « to be that one of the first pair that has the same 
parity as wu. Then $(#+wu).$(w@—u)=(Vt+y).% (v—y), where 
(+ u)/2, ete. are integral. Applying the lemma we have 
(w+ u/2=a8, (e©—w/2=98, W+Y/2=ay (wy /2= AS 
whence =a8+y6, y=ay— fd, w= aB—y5, v=ay+ B86, which 
gives a complete solution of the problem to find a sum of two 
squares equal to the sum of other two squares. These values of « 
and y give 22+ y°= (a8 + 7S) + (ay — Bo? = (a + &) (B+ 7"), and 
we easily see that § is the only one of the four that can vanish, 
so that only the first bracket can be unity. This requires that 
=u, y=v, so that the squares are the same. This is the well- 
known theorem that if a number can be expressed in more than 
one way as the sum of two squares, it is composite bie 
We can deduce the solution of a+ y2=2(w+). For it is 
clear that « and y are either both even or both odd, so that 
E=(24+y)/2 and »=(a—y)/2 are integral. Substituting for 
# and y, the equation reduces to the form just discussed, 
B+ Pa=wtv 
* The more general case ?-+ Ny?=u?+ Nv? can be treated in the same way by the 
extension of the lemma. The equation «3+ y®=u+v* was solved by Bachet (in his 
Diophantus). See also Fermat’s note (p. 133). The equation xt +yt=ut+v% also 
can be solved generally, one solution being 
2=12231, y=2903, w=10381, v= 10203. 
