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@. Case, Z. =. , V. V. Therefore, and C. , Not possible since otherwise In this case, as Y = Y and Z = Z, then U = (U , 0) with The following cases may arise. ? Case Y = (1, 0, V , 1) and Z = (1, U , 1) In this case, V is a palindrome because Z is a palindrome, and U is a palindrome because of Z . Moreover, ) = (U, 0) and (0, V ) = (V, 0). By Lemma 4, there are , ? 0 such that U = (0) and V =, p.1, 2001.

R. =. Hence, ) with + = n ? 2i ? 3 and , < i ? 2. Therefore) with + 1 >, p.1, 2001.

?. Case and Y. , By contradiction, we assume that both C1 and C2 belong to S 3 . Therefore, U = (1 j ) and A = (1 j , 0, 1) for some j, j > 0. As V is palindrome, then (1 j , 0, 1, 0 i?1 , B) = (B, 0 i?1 , 1, 0, 1 j ) As Z is palindrome, then (B, 1 From the first equality, we obtain that * B = (1 j ) and i = 2 or * B = (1, 0, 1 j ) or j ) where B = B . From the second equality, we obtain: * B = (1 j ) and i = 2 or * B = (1 j+1 , 0) or, B ) where B = B . Combining the two sets of equalities, we can have only the following cases. * B = (1 j ), j = j , and i = 2. In this case, Z = Z = ) and therefore, C1 = C2, a contradiction, 2001.

|. >. If and . |y-|, we define X such that X = (Y , X ) By plugging Y into (Y, X) = (X, Y ), we obtain, Therefore, X = (1 m ) for some > 0 and X = (Y, 1 m ), pp.1-1

?. Case and Y. , i?1 , 1, 0, A) and Z = (B, 1) with U = (A, 0 i?1 , B)