# Automorphic Forms, Representations and L-Functions, Part 2 by Oregon State University, 1977 Symposium in Pure Mathematics

By Oregon State University, 1977 Symposium in Pure Mathematics

Includes sections on Automorphic representations and L-functions in addition to Arithmetical algebraic geometry and L-functions

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Extra info for Automorphic Forms, Representations and L-Functions, Part 2

Example text

We shall ﬁnd (them) as follows. The 21 4 by itself, there results 16. Divide (it) by the 8, there results 2. Subtract the 2 from the 8, there remains 6. Half of it is 3. The height will be 3. Next, subtract the 3 from the 8, there remains 5. Hence the hypotenuse will be 5 feet. There is a right-angled triangle of which the height and the base are together 17 feet and the hypotenuse is 13 feet; to ﬁnd the height and the base separately. We shall ﬁnd (them) as follows. The 13 by itself (gives) 169.

Y Ã ÃÃ ? × Y× ... Ó ×Ú | ÑÓÒ ´ µ Ö Ñ ÒÑ Ó ÙÒ ÑÓ Ò Ñ ÙÒ Ñ Ù Ó Ù ÙÓ — ÙÒ ÑÓ Ù Ù Ó ? ÙÓÙ ÙÓ? Ö ÑÓ×Ø Ò ÙÒ ÑÓ×Ø Ò ... Ø ØÖ ÛÒÓ ×Ó (Ð Ô Ò) Figure 6 The Arithmetica is not a work of theory but a collection of problems. Primarily, these problems are indeterminate and quadratic (or reducible to a quadratic problem), with solutions that must be positive and rational (not necessarily integral, as the modern use of the expression “Diophantine equation” would imply). The few glimpses of theory contained in the Arithmetica appear in conditions that may be necessary for obtaining a rational solution; Diophantus mentions such conditions at the outset, immediately following the statement.

As in the ﬁrst case, there are inﬁnitely many choices of m yielding a positive value of x. (3) If either a or c is a square, it is once again possible to reach a rational solution via an elementary method. It suﬃces to make the square term appear as part of the indeterminate square, as eliminating the square term from each side will yield an equality between two consecutive powers of x, thus a rational value for x. = (Ax + m)2 , we Indeed, let A2 x2 + bx + c = . Setting obtain m2 − c . x= b − 2mA Similarly, if ax2 + bx + C 2 = , we set = (mx + C)2 , thus ﬁnding that b − 2mC .

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