How do I do the following problem?Let P=(1,3) be a point on the elliptic curvey^2=x^3+8.  Compute 2P, 4P and?
444950
2P = (2,4)\n4P = (4,sqrt(72))
1036703
To find the point of intersection between two curves, put y1=y2 so\n\nLet\ny1 = x^2 + 3x +2\ny2 = 7x -2\n\ny1 = y2\n\nx^2 + 3x + 2 = 7x -2\n\nRearrange:\n\nx^2 + 3x + 2 - 7x + 2 = 0 --->  x^2 - 4x + 4 =0\n\nThis is a second order eq. to solve it\n\nx1,x2 = [-b +- Sqr(b^2 -4 a c )] / (2 a)\n\nwhere a = 1,  b = -4 and c = 4\n\nx1,x2 = [+4 +- Sqr( (-4)^2 - 4 (1) (4) ) ] / (2*1)\n\nx1,x2 = [+4 +- Sqr( 16 - 16) ] / 2\n\nx1,x2 = 4/2 = 2\n\nthen x1=x2= 2  that means there is only one point; to find the corresponding y substitute x in either of the eq.\n\ny = 7 (2) -2 = 12\n\nThen the point (2,12) is the point of tangency because a parabola and a line intersect with two point. Therefore, (2,12) is the tangency point..!
800348
b^2 - y^2 - 2yp - p^2\n=b^2-(y^2+2yp+p^2)\n=b^2-(y+p)^2\n=[b+y+p][b-(y+p)]\n=(b+y+p)(b-y-p)
714206
x^2-2x-8 is factored as (x-4)(x+2).\n\nSo that means that +4 and -2 are your boundary points.  \n\nWithout checking specific points it looks like your answer would be A, but you have to check points in each of the intervals.  \n\nIf you pick -10 for x, (-10)^2 - 2(-10) - 8 > 0\n100 + 20 - 8 > 0\n112 > 0 True so that means that the interval from negative infinity to -2 works.\n\nIf you pick 0 for x, (0)^2 - 2(0) -8 >0\n0 - 0 - 8 > 0\n-8 > 0 False\n\nIf you pick +10 for x, (+10)^2 - 2(+10) - 8 > 0\n100 - 20 - 8 > 0\n72 > 0 True so that means the interval from 4 to positive infinity works.\n\nSo after all that work and checking answer A seems to be the correct one.\n\nHope this helps.  Good Luck.  :)
771200
You are 100% right.\n(4P+11)(4P-11)\n=(4P)^2-(11)^2\n=16P^2-121
899635
Something is weird with the problem: you do not need the vertex. Three points are plenty to determine a parabola. Here's how:\n\n[(x-1)*(x+3)] / [(-1-1)*(-1+3)]  is zero at x=1 and x=-3, but one at x=-1\n\nSimilarly construct two others\n\n[(x+3) * (x+1)] / [(1+3) * (1+1)]  \n\nand \n\n[(x-1) * (x+1)] / [(-3-1) * (-3+1)] \n\nSo you have three parabolas, each one zero on two points and one on the third. Multiply each by the value which your parabola should have on the appropriate point and add. \n\nYou could also say a x*x + b *x +c are three unknowns and get three equations from the three prescribed values, but this is more tedious and less enlightening.
457571
the answer a:108\nbecause the intersection between the two curve ;-3and3\n[  x^2=-2x^2+27   the value of x are 3 and -3  ]\nyou will integrate from -3 to 3 (-2x^2 +27 - (x^2)) dx\n[-x^3+27x] from -3 to 3 \nsubtitute you will get 108
421674
y = (x mod 5) + 1.\n\nThough that does seem a bit like cheating :)
586333
First, FUNNYPLAYA12'S answer is a partial of what I got, and my formula looked like this:\n\n-8X^3 + 10X^2 - 2X = 0\n-----------------------------------\nFactor out one X and you get:\n\n-8X^2 + 10X - 2 = 0\n\nFactor this and FOIL-check\n\n(-4X + 1)\n(2X - 2)\n___________\n-8X^2 (8X + 2X) -2\n\n-8X^2 +10X - 2\n\nX = {1/4, 1)\n\n\nTo verify:\n\n-8(1) + 10(1) -2(1) = 0\n\n-8 + 10 -2 = 0; and finally:\n\n10 - 10 = 0, which is very true!\n----------------------------------------\n\n-8(1/4) + 10(1/4) -2(1/4) = 0\n\n-2 + 2.5 - 0.5 = 0\n\n-2.5 + 2.5 = 0, which is also very true!\n\n\nOTBPOC
103938
you have to solve p such that S=D=Q.\n\nfrom S, p=Q/10 -70\n\nput that instead of p into D equation, solve for Q\n\nthen use that Q to compute p=Q/10 -70
859279
This would be a special case of the Fermat-Catalan Conjecture.\n\nThis conjecture is that there are only a finite number of solutions of x^p + y^q = z^r for 1/p + 1/q + 1/r < 1.  Ten solutions are known, but all of them have the three exponents different (except for 1^p + 2^3 = 3^2) and one exponent equal to 2.\n\nYour conjecture would be the special case that there are no solutions with p > 2 and q = r.  It is already known that there are no solutions with p = q and r > 2 (Darmon and Merel, 1997).
1148686
(2y-8)(y-4)\n\n2y^2-8y-8y+32\n\n2y^2-8y^2+32
940136
Hahahahaha good luck on that one... it took Andrew Wiles seven years to prove :)\n\nFirst you need to prove the following theorem:\nSuppose E is an elliptic curve with integer coefficients in global minimal form. If it has discriminant Δ a product of primes p with exponents δp, and conductor N a product of primes p with exponents np, and if E is a modular curve (now known to be true), then we can perform a level descent modulo primes ℓ dividing one of the exponents δp of a prime dividing the discriminant. If pδp is an odd prime power factor of Δ and if p divides N only once, then we can descend to another elliptic curve E' with conductor N' = N/p, such that the L-series coefficients for L(s, E) and those for L(s, E') are congruent modulo ℓ.\n\nOnce you're done with that, prove this:\nAny elliptic curve over Q can be obtained via a rational map with integer coefficients from the classical modular curve\n\n    X0(N)\n\nfor some integer N; this is a curve with integer coefficients with an explicit definition. This mapping is called a modular parametrization of level N. If N is the smallest integer for which such a parametrization can be found (which by the modularity theorem itself is now known to be a number called the conductor), then the parametrization may be defined in terms of a mapping generated by a particular kind of modular form of weight two and level N, a normalized newform with integer q-expansion, followed if need be by an isogeny.\n\nFrom another point of view, given an elliptic curve E over Q we may define a corresponding L-series. The L-series is a Dirichlet series which we may write\n\n    L(s, E) = &#92;sum_{n=1}^&#92;infty &#92;frac{c_n}{n^s}.\n\nWe can take the same coefficients, and use them to define a function in powers of q\n\n    f(q, E) = &#92;sum_{n=1}^&#92;infty c_n q^n.\n\nIf we make the substitution q = exp(2πiτ), then the series becomes a Fourier series, and so the coefficients are sometimes called "q-series coefficients", but other times "Fourier coefficients". The function obtained in this way, remarkably, is a cusp form of weight two and level N and is also an eigenform (an eigenvector of all Hecke operators); this is the Hasse-Weil conjecture, which follows from the modularity theorem.\n\nSome modular forms of level two, in turn, correspond to holomorphic differentials for an elliptic curve. The Jacobian of the modular curve can (up to isogeny) be written as a product of irreducible abelian varieties, corresponding to Hecke eigenforms of weight 2. The 1-dimensional factors are elliptic curves (there can also be higher dimensional factors, so not all Hecke eigenforms correspond to rational elliptic curves). The curve we obtain by finding the corresponding cusp form, and then constructing a curve from it, is isogenous to the original curve (but not in general isomorphic to it).\n\n\nThat should get you started.
865971
p(x)=-3x^2 (x-1) (x+1/3) (x+2)^3\n\n1) what are the roots and multiplicities? \nroots - multiplicity\n0 .............2\n1..............1\n-1/3 .........1\n-2 ............3\nbasically the multiplicity of a root A is the exponent of the factor (x-A)\n.
1177326
1) q^3/(5p^8)\n\n2) 1/(-2y)^4 = 1/[(-2)^4(y)^4]\n= 1/(16y^4)
930651
Distributive property:\n8p+8+3p=-14\n\nCombine like terms:\n11p+8=-14\n\nSubtract 8 on both sides:\n11p=-22\np=-2\n\nCheck:\n8(-2+1)+3(-2)=-14\n8(-1)+3(-2)=-14\n-8-6=-14\n-14=-14
747036
0.12 \nP(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - P(A)P(B|A) = 0.1 + 0.1 - 0.1(0.8)