Since we are told b divides 2a², then let k=2a²/b, so k is an integer and kb=2a². It is to prove a²+b is not a perfect square. To show this, it can be shown if it did, then it leads to a contradiction.
Suppose a²+b=x² for some integer x. Then subtract a² from both sides and multiply by k to get kb=k(x²-a²)=k(x-a)(x+a). Substitute
kb=2a² to get k(x-a)(x+a)=2a².
Now note g1=gcd(x-a,a) and g2=gcd(x+a,a) are equal, for positive integers x and a. To show this, note g1 divides x+a, since if g1 divides x-a and a, then it must divide x and thus x+a. Thus, g1 is a common divisor of x+a and a, so g1 divides g2. A similar proof can show g2 divides g1, thus g1=g2. Let g be this common gcd.
Then (x-a)/g and a are relatively prime, so ((x-a)/g)² and a² are relatively prime, and thus (x-a)/g and a² are relatively prime. Similarly, (x+a)/g and a² are relatively prime.
Next is to show that (x-a)/g and (x+a)/g are relatively prime to 2a². Let p be the largest power of 2 which divides a. Assume 2p divides (x-a), thus 2p must divide x+a=(x-a)+2a since 2p divides 2a. Thus, 2p*2p=4p² divides 2a², but this
contradicts p being the largest power of 2 in a. (It can be shown similarly that assuming 2p divides (x-a) leads to a contradiction.)
Thus, p is the largest power of 2 in x+a and x-a. This shows (x+a)/g and (x-a)/g must be odd since p divides g. Thus, (x-a)/g and (x+a)/g are relatively prime to 2a².
If (x-a)/g>1, or (x-a)/g>1 then x-a or x+a would contain a factor ((x-a)/g or (x+a)/g) that 2a² would not. Since x>a then (x-a)/g>=1 and (x+a)/g>=1, so (x-a)/g=(x+a)/g=1, and (x-a)/g=(x+a)/g implies x-a=x+a.
However, x-a=x+a implies -a=a, 2a=0, a=0, a contradiction since a and b must be positive. So if one has (a,b) such that b divides 2a², then a²+b isn't a perfect square.
Posted by Gamer
on 2008-05-07 17:22:09