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Extra info for A solution to a problem of Fermat, on two numbers of which the sum is a square and the sum of their squares is a biquadrate, inspired by the Illustrious La Grange

Example text

Vn/Vn+ 1 9 Consider the case when ~ ~ 3, The map raising to the g-th power gives an is omor phis m Vn/Vn+ I ~ Vn+i/Vn+ 2 9 Let U 1 be a closed subgroup of V 1 and let U n = U I fIV n. We have an embedding U1/U 2 9 Vl/V 2 ~ M/L~ . We may refer to U1/U 2 as the a s s o c i a t e d vector space to U 1 at level 1, or as the tangent space to U 1. It has dimension g 4. 47 Lemma 2. L e t g be odd, and let U 1 be a closed subgroup of V 1 whose associated vector s p a c e at level 1 has dimension 4. Then U 1 = V 1.

There e x i s t s a continuous density function 4(0) on [0, ~z] determining the distribution o[ angles O(tp, p). This means that the primes p such that 0(tp, p) lies in a given interval [01, 02] have a density, namely 02 J 4(0) dO. 01 Changing variables, with ~: = cos 0 shows that the condition ST is equivalent with: ST: There e x i s t s a continuous function g(f) = 6(o) which is equal to 0 outside the interval [-1, 1], such that the primes p for which ~:(tp, p) lies in an interval [~:1' ~:2] have a density, given by f2 f g(~:)dr.

From t h i s we g e t the s i m i l a r lemma: Lemma 3. L e t g = 2. L e t U 1 be a closed subgroup of V 1. If U 2 / U 3 = V 2 / V 3 then U 2 = V 2. If furthermore U 1 / U 3 = V t / V 3, in other words, if the reduction of U 1 rood 8 contains I+2M 2(mod8), then U I = V 1 . Proof. Clear. Abelianization We l e t G be our u s u a l group with r e p r e s e n t a t i o n p. T h e f i e l d K of which G is the G a l o i s group c o n t a i n s a l l the roots of unity, and t h i s c y c l o t o m i c field is the maximal a b e l i a n e x t e n s i o n of the r a t i o n a l s .

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