3-Sasakian Geometry, Nilpotent Orbits, and Exceptional by Boyer Ch. P.

By Boyer Ch. P.

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The latter translates into π/4 11 4 tan−1 (1/5) − tan−1 (1/239). For more details, see J. Conway and R. Guy, The Book of Numbers, Springer, 1996. ) Derive the last formula using trigonometry. (Let tan θ 1/5 and use the double angle formula for the tangent to work out tan(2θ) 5/12 and tan(4θ) 120/119. Notice that this differs from tan(π/4) 1 by 1/119. ) 15. ♦ Given a basis {v, w} in R2 , the set L {kv + lw | k, l ∈ Z} is called a lattice in R2 with generators v and w. Generalize the concept of fundamental parallelogram (from Z2 ) to L, and show that every possible fundamental parallelogram of a lattice L has the same area.

If a, b, c are the rational side lengths of a right triangle with congruent area n, then multiplying the two equations ((a ±b)/2)2 (c/2)2 ±n together, we get ((a2 −b2 )/4)2 4 2 (c/2) − n . We see that the equation u4 − n2 v2 has a rational 2 2 solution u c/2 and v (a − b )/4. Setting x u2 (c/2)2 and y uv (a2 − b2 )c/8, we obtain that (x, y) is a rational point on the elliptic curve in the Weierstrass form y2 x3 − n2 x. ♣ The theory of elliptic curves displays one of the most beautiful interplays between number theory, algebra, and geometry.

27 Rationality, Elliptic Curves, and Fermat’s Last Theorem of the equation by c2 d 2 , we obtain a 2 d 2 + b2 c 2 c2 d 2 . This immediately tells us that c2 divides a2 d 2 , and d 2 divides b2 c2 . Since a, c and b, d have no common divisors, c2 and d 2 divide each other, and so they must be equal. The equation above reduces to a 2 + b2 c2 . We see that the question of rationality is equivalent to finding positive integer solutions of the Pythagorean equation above. A solution (a, b, c) with a, b, c ∈ N is called a Pythagorean triple.

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