244 CHAPTER 7 NUMBER THEORY
so we can conclude that n > m. That certainly helps. Let's consider one example, with,
say, m = 1, n = 2. Our diophantine equation is now
.x2+i =.x2i·
Factoring quickly establishes that there is no solution, for adding 1 to both sides yields
(.x2 - 1)(i - 1) = 1,
which has no positive integer solutions. But factoring won't work (at least not in anobvious way) for other cases. For example, let's try m = 3,n = 4. We now have
(7)The first thing to try is parity analysis. A quick perusal of the four cases shows that theonly possibility is that both x and y must be even. So let's write them as x = 2a,y = 2b.
Our equation now becomes, after some simplifying,(a^2 +b^2 )^3 =4(ab)^4.
Ponder parity once more. Certainly a and b cannot be of opposite parity. But they
cannot both be odd either, for in that case the left-hand side will be the cube of an
even number, which makes it a multiple of 8. However, the right-hand side is equal to4 times the fourth power of an odd number, a contradiction. Therefore a and b must
both be even. Writing a = 2u, b = 2v transforms our equation into
(u^2 + i)^3 = 16(uv)^4.
Let's try the kind of analysis as before. Once again u and v must have the same
parity, and once again, they cannot both be odd. If they were odd, the right-hand sidewould equal 16 times an odd number; in other words, 24 is the highest power of 2 that
divides it. But the left-hand side is the cube of an even number, which means that thehighest power of 2 that divides it will be 23 or 26 or 29 , etc. Once again we have a
contradiction, which forces u, v to both be even, etc.
It appears that we can produce an infinite chain of arguments showing that the
variables can be successively divided by 2, yet still be even! This is an impossibility,
for no finite integer has this property. But let's avoid the murkiness of infinity by using
the extreme principle. Return to equation (7). Let r, s be the greatest exponents of 2that divide x, y respectively. Then we can write x = 2r a, y = 2s b, where a and b are
odd, and we know that rand s are both positive. There are two cases:
- Without loss of generality, assume that r < s. Then (7) becomes
( 22 r a (^2) + 2 (^2) Sb^2 )^3 = 2 (^4) r+ (^4) sa (^4) b (^4) ,
and after dividing by 26 r, we get
(a^2 + 2^2 r-^2 sb^2 )^3 = 2^4 s-^2 r a^4 b^4.
Notice that the exponent 4 s - 2r is positive, making the right-hand side even.
But the left-hand side is the cube of an odd number, which is odd. This is an
impossibility; there can be no solutions.