By B.M.M. de Weger
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Extra resources for Algorithms for Diophantine Equations
I i i=1 We assume that n > 2 . 2, but the method of this section works also for n = 2 . In fact, it is in this case essentially the same method. Let Let C n . 0 be a constant (we will explain its use later). We define the be a large enough integer, that is of the order of magnitude of g e N approximation lattice G X by the matrix ( g | . o | . B = | o | g | [gWCWy ] ... , b are a basis of the lattice. Then G 1 n n n-1 is a sublattice of Z of determinant g W[gWCWy ] , which is of size C . , gWx , ~L )T , S x Wb = i i 9 1 n-1 0 i=1 are integers, and n S x W[gWCWy ] .
We have for the linear form L (cf. 4)), ord (L) > c + c Wm , p 1 2 j where c , c are small constants, and m is one of the variables. 1 2 j Moreover, the variables are bounded by a large constant N , that is 0 m m+1 explicitly known. We take m such that p is at least of size N , so 0 * that the lower bound for the shortest nonzero vector in G (or G ) is rmWN0 . 2). Therefore, + c Wm < m , 2 j so that we find a new upper bound for m , that is of the size of m , which j is about log N / log p . We repeat this procedure for all the m , in 0 j order to obtain a reduced upper bound for H .
6). 8. |x | i for different i . 24) holds. Then 1 & ( ~ n )* . X < -----Wlog gWCWc/ |L|- S |x | d 7 9 i 08 i=1 Proof. 7. 24) |L| > n (|L|~ ) S |x | /gWC > 0 . 1). 24). 25). In case |L| < C we have an upper bound for the 0 length of the vector x . We compute all lattice points satisfying this bound We proceed as follows. 1). 1) corresponds to an extremely short vector in an appropriate approximation lattice. Since we can actually prove by computations that such short vectors do not exist, it follows that such large solutions do not exist.
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