By Euler L.
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Extra resources for An illustration of a paradox about the idoneal, or suitable, numbers
Proof. In fact, the above inequalities hold true for the Taylor coefficients at 0 of any function which is holomorphic in an open neighborhood of the unit disk |z| ≤ 1, satisfies |f (z)| ≤ 1 for |z| = 1 and has real Taylor coefficients cj . Indeed, setting p(z) = x0 + x1 z + · · · + xn z n , we have n 1 (c0 xj + · · · + cj x0 ) = 2π j=0 2π |f (z)p(z)|2 dt 2 1 ≤ 2π 0 n 2π x2j . 2 0 (z = eit ) |p(z)| dt = j=0 The second assertion is obtained using the Cauchy-Schwartz inequality and the first one: ±xt Ax ≤ |x| |Ax| ≤ |x|2 .
3. 2. The image of φ is contained in G2p , p|∆E or p=∞ where ∆E = (a − b)2 (a − c)2 (b − c)2 is the discriminant of E. In particular, it is finite. Proof. Let P ∈ E(Q), P = 0, say P = (x, y). Let p be a prime number, and let φ(P ) = (uQ∗ 2 , vQ∗ 2 ). We have to show that ordp (u) and ordp (v) are both even. For this let pn be the exact power of p in the prime decomposition of x. If n < 0, then x = 0, a, b and we can take u = x − a and v = x − b. Since a, b, c are integral pn is also the exact power of p in x − a, x − b and x − c.
Thus C ∗ : x + y = xy. The intersection points of C and C ∗ are ρ = 1+ 2 −3 and its complex conjugate. If we take for the G used in the preceding proof G = (ρx − ρy)(ρx − ρy) = x2 − xy + y 2 , 34 PART 2. HEIGHTS ON ELLIPTIC CURVES then G∗ = x2 − xy + y 2 , and 1 1 γs (1, 1 − z) = |z| 2 |1 − z| 2 |z 2 − z + 1| |z||1 − z| 2s . √ This is the function we actually used in our original proof with s = 1/4 5. 2 Heights on projective space For a point P in Pn , say with projective coordinates [x0 : · · · : xn ] in a number field K, we define its height HK (P ) relative to K and its absolute height H(P ) by HK (P ) = v∈PK H(P ) = HK (P )1/[K:Q] .