Note that for the same reasons as in the last post, x and y must be nonnegative and similarly, x must be twice a perfect square.
If the same method is applied for the end of the root: sqrt(x+sqrt(2x))=z²
(sqrt(x/2)+1)*2sqrt(x/2)=z²
Since 2x has to be a perfect square, x must be even, so x/2 is also a perfect square. Thus for sqrt(x/2)>1, the terms are relatively prime. This means sqrt(x/2) needs to be a perfect square ie x is of the form 2b^4.
Also, sqrt(x/2)+1 needs to be 2a² for some integer a, in order the product to be a perfect square. Subtract 1 from both sides, and one can see sqrt(x/2) is part of a sophie square chain. If b=fourth root(x/2), then 2a²-1=b² We have already seen that x is twice a perfect fourth power, but b needs to be one of the ends of a sophie square chain, thus b=1, 7, 41... resulting in x=2, 4802, 5651522.. and so only two numbers work here for x < 180,000. Substituting these in, x=4802, gives z=70, but 4802+70=4872 is not a perfect square. x=2 is a solution, since it gives z=2, and one can see each successive part is 2, thus y=2 for x=2.
If sqrt(x/2)<=1, then x=2 (which is a solution), and x=1, and x=0 are the only integers left to be tested. x=1 is not a solution but x=0 is. So the only x which satisfy the equation and x<180,000 are x=0 and x=2.
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Posted by Gamer
on 2007-02-20 22:58:02 |
Solution for B (using same method)
