7.4 DIOPHANTINE EQUATIONS 245• Now assume that r = s. Then (7) becomes
(a^2 +b^2 )^3 = 22 r a^4 b^4. (8)
It is true that both sides are even, but a more subtle analysis will yield a con
tradiction. Since a and b are both odd, a^2 == b^2 == 1 (mod 4f' so a^2 + b^2 == 2(mod 4), which means that (^21) 1Ia^2 +b^2. Consequently, 23 11(a +b^2 )^3. On the
other hand, 22 rl 122 r a^4 b^4 , where r is a positive integer. It is impossible for 2r = 3,
so the left-hand and right-hand sides of equation (8) have different exponentsof 2 in the PPFs, an impossibility.
We are finally ready to tackle the general case. It seems as though there may be
no solutions, but let's keep an open mind.
Consider the equation (.x2 + y^2 )m = (xy t. We know that n > m and that both x and
yare even (using the same parity argument as before). Let 2 rllx,2slly. We consider the
two cases:
1. Without loss of generality, assume that r < s. Then 22 rmII (.x2 + y^2 )m and
2 nr+nsll(xy)n. This means that 2rm = nr + ns, which is impossible, since m
is strictly less than n.
2. Assume that r = s. Then we can write x = 2 r a,y = 2 rb, where a and b are both
odd. Thus(x^2 + i)m = 22 rm(a^2 + b^2 )m,
where a^2 +b^2 ==^2 (mod 4) and consequently 22 rm+mll (x^2 +y^2 )m. Since 22 nrll(xy)n,
we equate2rm+m = 2rn,
and surprisingly, now, this equation has solutions. For example, if m = 6,
then r = 1 and n = 9 work. That doesn't mean that the original equation has
solutions, but we certainly cannot rule out this possibility.
Now what? It looks like we need to investigate more cases. But first, let's thinkabout other primes. In our parity analysis, could we have replaced 2 with an arbitrary
prime p? In case 1 above, yes: Pick any prime p and let pUllx,pVlly. Now, if we
assume that u < v, we can conclude that p^2 umll (x^2 + y^2 )m and pnu+nvil (xy)n, and this
is impossible because n > m. What can we conclude? Well, if it is impossible that u
and v be different, no matter what the prime is, then the only possibility is that u and vare always equal, for every prime. That means that x and yare equal!
In other words, we have shown that there are no solutions, except for the possible