From: caebi@infomaniak.ch (Christian Aebi) Newsgroups: sci.math Subject: Solving x^4+x^3+x^2+x+1=y^2 for integers Date: 12 Jan 1999 08:41:03 -0500 I beleive I have a method, but I'd like to know : "Is there an easier one ??" This is the way I solved it : 1)I factored x^4+x^3+x^2+x+1 in Q(sqrt(5)) into two conjugate polynomials of degree 2 that I'll call A(x) and A*(x). The algebraic integers of Q(sqrt(5)) will be called R(sqrt(5)) 2) Each expression A and A* are prime to another if 2 is prime in R(sqrt(5)). 3) I proved 2 is prime in R(sqrt(5)), using the norm N(x) and working mod(8). 4) Each expression A and A* must then be squares since R(sqrt(5)) is a factorial ring (That, I have not yet proved). 5)There are two kinds of algbraic integers in R(sqrt(5)). Those of the form a+b*sqrt(5) with a and b belonging to Z and those like (a+b*sqrt(5))/2 with a==b==1 mod(2) Substituting y by an algebraic integer of the latter forms, then: working modulo 4 in the first case , and working out second degree equations in the second case gives the solutions x=0 and y = +/- 1 or x=3 and y = +/- 11 . Hoping it's not looking too messy ! Christian ============================================================================== From: kubo@turing.math.brown.edu (Tal Kubo) Newsgroups: sci.math Subject: Re: Solving x^4+x^3+x^2+x+1=y^2 for integers Date: 15 Jan 1999 16:09:28 -0500 Dave Rusin wrote: >John Robertson wrote: >>Note that Y^2 = X^4 + X^3 + X^2 + X + 1 is an elliptic curve. Use the >>methods in J. W. S. Cassels, Lectures on Elliptic Curves, Chapter 8, >>section (iii), to put this equation into Weierstrass form. I got y^2 >>= x^3 - 5x^2 - 45x + 25, where X = (y-x-5)/4x and Y = (x/8) - X^2 - >>X/2 - 3/8. >>Determine that all integral points on this form of the curve >>are [x, y] = [-5, 0], [-1, +/-8], [0, +/-5], [11, +/-16], [15, +/-40], >>or [175, +/-2280]. Map back to X and Y, and you'll see that the only >>integer points on your original equation are those that you list. > >It's a non-trivial task to know that a given set of integers is complete, >but even assuming the eleven points are the only integrals points on >y^2=x^3-5x^2-45x+25, you still don't really know what the integral points >on the original curve are. Your transformation between the curves does >not (necessarily) pair off the integral points on them. in this case we see that x = 8Y + 8X^2 + 4X + 3 y = 4X*x + x + 5 so if X,Y are both integers then so are x,y. But for this particular equation there is an elementary solution; the polynomial in X is squeezed between f(X)^2 and (1+f(X))^2 for an integer-valued polynomial f(X). ============================================================================== [Note: This is correct. Kubo's suggestion is valid for equations of the form Y^2 = Q(X) where Q is a monic quartic. When Q is not monic, there are often transformations as Robertson suggested but they will involve rational functions, not polynomials, so Rusin's caveats apply. If Q(X) is monic (or has lead coefficient a perfect square), then the square roots of Q(X) can be computed as F(X) + R(X) where F is a quadratic polynomial whose (rational) coefficients are determined from those of Q, and R(X) is an error term which can be estimated by Taylor's formula and which can thus be shown to be less than any pre-determined epsilon, once X is sufficiently large. In particular, if X and Y are integers and Y^2 = Q(X), then (+-Y) - F(X) is at once rational (with bounded denominator) and small, hence zero. Thus Y = +- F(X). Thus except for small solutions, all other solutions will require Q(X) = F(X)^2; there are at most three such X. All these bounds may be found effectively. -- djr] ============================================================================== From: pete@bignode.southern.co.nz (Pete Moore) Subject: Re: Solving x^4+x^3+x^2+x+1=y^2 for integers Newsgroups: sci.math Date: 16 Jan 99 22:42:04 +1200 John Robertson (jpr2718@aol.com) wrote: >I don't know if this approach is simpler, but it's different. > >Note that Y^2 = X^4 + X^3 + X^2 + X + 1 is an elliptic curve. Use the >methods in J. W. S. Cassels, Lectures on Elliptic Curves, Chapter 8, >section (iii), to put this equation into Weierstrass form. I got y^2 >= x^3 - 5x^2 - 45x + 25, where X = (y-x-5)/4x and Y = (x/8) - X^2 - >X/2 - 3/8. > >Use the Pari function ellglobalred to determine that the conductor of >this curve is 200. [etc] Eeek! There's a much simpler way: Write the equation as (2y)^2 = 4x^4 + 4x^3 + 4x^2 + 4x + 1 And note that the right side is greater than (2x^2+x)^2, and is also equal to (2x^2+x+1)^2 - (x^2-2x-3) and hence is less than (2x^2+x+1)^2, except possibly for those values of x for which x^2-2x-3 is negative (i.e. -1<=x<=3) - for other values of x, the right side lies between two successive squares, and so cannot be a square. Considering the values -1<=x<=3 individually, we find that the only solutions are (x,y) = (-1,1), (0,1), (3,11) -- +---------------- pete@bignode.southern.co.nz ----------------+ The effort to understand the universe is one of the very few things that lifts human life above the level of farce, and gives it some of the grace of tragedy - Steven Weinberg