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... While it is not difficult to constructs for any n ^ 3* a magic square of order n s it seems formidable to determine Q 0 (n) or 60(n) for n ^ 6 (see [1] and [2]). In [4], it is shown that 60(4) = 48 and in [5] that <50(5) = 3600. In this note, we shall show that, given an odd-order pandiagonal magic ...
... While it is not difficult to constructs for any n ^ 3* a magic square of order n s it seems formidable to determine Q 0 (n) or 60(n) for n ^ 6 (see [1] and [2]). In [4], it is shown that 60(4) = 48 and in [5] that <50(5) = 3600. In this note, we shall show that, given an odd-order pandiagonal magic ...
Squares in arithmetic progressions and infinitely many primes
... Another proof of Theorem 2. Fix an integer M > 6/δ . Let B(M ) be the total number of rational numbers x and integer 6-tuples b1 = 0 < b2 < . . . < b6 ≤ M − 1 for which (x + b1 )(x + b2 ) · · · (x + b6 ) is the square of a rational number. Faltings’ theorem implies that B(M ) is some finite number, ...
... Another proof of Theorem 2. Fix an integer M > 6/δ . Let B(M ) be the total number of rational numbers x and integer 6-tuples b1 = 0 < b2 < . . . < b6 ≤ M − 1 for which (x + b1 )(x + b2 ) · · · (x + b6 ) is the square of a rational number. Faltings’ theorem implies that B(M ) is some finite number, ...
ppt - Marric.us
... up so that one number is to its left Step #3: Count how many places you bounce the decimal point Step #4: Re-write in the form M x 10n ...
... up so that one number is to its left Step #3: Count how many places you bounce the decimal point Step #4: Re-write in the form M x 10n ...
