From: rusin@shavano.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Triangle Areas, Rational Squares, and Galois Theory. Date: 4 Jun 1999 18:12:02 GMT Keywords: Yet more elliptic surfaces While thinking about one of the problems which appeared recently in this newsgroup, I realized there was a connection with some other threads (as well as with some other unsolved problems). One way to develop the connection uses some Galois theory so I thought I'd post in detail the development, in case anyone wants to see what good Galois theory can be! Unfortunately I don't actually solve the original problems... **************** First the newsgroup threads. In article <3746e41a.3569655@news.gte.net>, rbergmann@ariel.com (Rick Bergmann) asked about Rational Rectangles In a Circle: >Given four rectangles inscribed in a circle, with areas >A1,A2,A3,A4 such that A1+A2=A3 and A1-A2=A4, is it possible >for the edges of the rectangles and the diameter of the >circle to all have rational lengths? We may scale the circle to have radius 1 and halve the equations to see we are looking for right triangles with rational sides and hypotenuse 1 whose areas add: A1+A2=A3, A2+A4=A1. Thus the areas A4, A1, A3 are in arithmetic progression. This, perhaps, led to another post: In article <374a3295.58926259@news.gte.net> sestnick@ariel.com (Steve Estnick) asked about Triangle Areas in Progression: >Can the areas of three rational right triangles on the same >hypotenuse be in arithmetic progression? So Bergmann's question is at least as hard as Estnick's. Actually Bergmann's question implies another challenge: can we even find three rational right triangles with hypotenuse 1 whose areas sum (i.e. A1+A2=A3) ? In article <19990526233527.23178.00005814@ng-cm1.aol.com> jpr2718@aol.com (Jpr2718) "John" followed up Estnick's post: >This is equivalent to the question as to whether there can be two 3-term >arithmetic progressions with the same differnce between terms, so that every >entry in both progressions is an integer square, and the average of the two >central terms is also a square. > >This is an open problem. It's interesting that the geometric problems lead to progressions of squares, because that theme is picked up in another thread (prompted by this?) In article <7j3ad8$3t1@news.acns.nwu.edu>, lerma@math.nwu.edu (Miguel A. Lerma) asked about powers in arithmetic progression : > It is possible to find many arithmetic progressions made up of three > perfect squares, such as 1, 25, 49, but are there any arithmetic > progressions made up of four or more squares (with positive difference)? (The answer to which is "no", as proved by Fermat.) **************** What I'd like to do in this post is to show how the questions about areas of rational right triangles lead to squares in progressions. It's true that the connections between these themes can be demonstrated by displaying some completely unmotivated algebraic substitutions. However, I found that the search for these formulas is sped up, and perhaps made clearer, by the use of symmetry arguments which can be used as examples of elementary Galois theory. The main result is this: THEOREM. A number r is the area of a rational right triangle with hypotenuse = 1 iff both 1-4r and 1+4r are rational squares. Of course, 1 is also a rational square, so we have three squares in arithmetic progression. We may scale this to express it in integers: COROLLARY. The areas of integer right triangles with hypotenuse = n are in one-to-one correspondence with the 3-term arithmetic progressions of squares n'^2 < n^2 < n"^2 (the common differences are 4 times the areas) So Lerma's example shows there is an integer right triangle with hypotenuse 5 and area 6. (I'll bet you knew that...) Fermat's theorem precludes certain kinds of pairs of integer right triangles having the same area. This material is hardly new, and indeed it goes much deeper: arranging for three things to be simultaneous squares amounts to doubling points on elliptic curves, and then the squares are to be in arithmetic progression, this refers to particular, well-studied elliptic curves. See e.g. Jpr2718's old sci.math post at http://forum.swarthmore.edu/epigone/sci.math/zharjorsal Or look up the "congruent number problem" (when is there a rational right triangle with a given area?), e.g. at 96/congruent.nos I just thought it would be of interest to use this as an example of Galois theory in action, helping us find the necessary formulas by symmetry. **************** Before I prove the theorem, I'd like to discuss the Diophantine problems this theorem then provides. Estnick's problem may now be stated, Q1. Are there distinct positive rational numbers a and b for which all of { 1 +- a, 1 +- b, 1 +- (a+b)/2 } are squares? As I noted, Bergmann's problem implies this and another "easier" one: Q2. Are there positive rational numbers a and b for which all of { 1 +- a, 1 +- b, 1 +- (a+b) } are squares? while Bergmann's original problem is equivalent to Q3. Are there positive rational numbers a and b for which all of { 1 +- a, 1 +- b, 1 +- a +- b } are squares? (By Fermat's theorem, a and b are necessarily distinct in these last two problems.) We may remove the references to a and b and write these as sets of nonlinear equations. For example, Q2 may be stated Q2. Are there nonzero rational numbers x, y, z for which 2-x^2, 2-y^2, 2-z^2 are all squares, and x^2+y^2=z^2+1 This is in reality a set of four equations in six variables, and thus describes an algebraic surface, on which we are asked to find a rational point. The other two problems also describe algebraic surfaces, the last surface being a cover of the other two and therefore less likely to have rational points. As Jpr2718 mentioned in his post, there is a connection to another open question: is there a 3x3 magic square each of whose entries is a perfect square? See http://www.seanet.com/~ksbrown/inumber.htm for more information; in particular, it is shown there that this is precisely the same as Q3, i.e. as Bergmann's question. (As I read Jpr's post, it seems he is claiming the magic-square problem to be equivalent to Estnick's problem, but I don't see that these are the same.) For comparison I'd like to mention a couple of other classic questions which sound almost the same and whose answer is still not known. The question of whether or not there is a "rational box" (sides, face diagonals, body diagonal all rational) also asks for a rational point on a surface; it can be written as Q4. Are there nonzero rational numbers x, y, z for which 1-x^2, 1-y^2, 1-z^2 are all squares, and x^2+y^2+z^2=1 I could also mention Euler's extension of the Fermat conjecture: are there three fourth powers which sum to another fourth power? This can be stated Q5. Are there nonzero rational numbers x, y, z, for which x, y, z are all squares, and x^2+y^2+z^2=1 One might imagine by comparing to the other problems that this one is easier, and indeed it has been solved (in the affirmative, by Elkies). For some perspective I point out that the answer to Q5 was obtained not all that long ago, and the solutions are not "easy", exactly (e.g. there appears not to be any parameterized family of theme). There are a number of other problems which can be cast in a similar framework -- they ask for rational points on a surface -- and for which no solution is known and no proof is known that there is no solution. For many of them it's possible to get really close to a solution, e.g. we only need to find two rational numbers making a certain polynomial in those two variables be a perfect rational square. If those polynomials had degree four or less in one of the variables we'd have a fighting chance of finding solutions using tools from elliptic curves (that's what Elkies did for Q5) but the polynomials are just too complicated for that. (I believe I once showed how to reduce the Euler problem for exponent 5 to just such a Diophantine equation, for example.) **************** Now I'd like to prove the main theorem connecting rational right triangles and sets of rational squares. A rational right triangle with sides of lengths a and b and hypotenuse 1 determines a point (a,b) in the first quadrant of the plane and lying on the unit circle. Actually it determines two such points, (a,b) and (b,a), which can be obtained from each other by using the "swap" reflection of the plane. If we don't mind leaving the first quadrant, we can find two other points from these two using the "spin" 180-degree rotation around the origin, another automorphism of the plane of order 2 (commuting with "swap"). As is well known, we may parameterize the unit circle rationally by drawing lines through (-1,0): we may draw a unique line joining any other point on the circle to this one, and conversely as we range through the lines through (-1,0) with every real slope t, we find each one meets the unit circle in precisely one more point, namely (x,y) = ( (1-t^2)/(1+t^2), (2t)/(1+t^2) ) The area of the triangle xy/2 is then given as a rational function of t: A(t) = t(1-t^2)/(1+t^2)^2 What's of interest to us is that the _rational_ points on the circle clearly correspond to _rational_ slopes t. Thus we see that a number r is the area of a rational right triangle with hypotenuse 1 iff there is a rational solution to the equation A(t) = r. So the rest of this section addresses the question of when there is a rational t with A(t)=r. The point of the geometric motions discussed earlier is the conclusion that if A(t) = r has one solution, it will have several others. It follows easily from high-school geometry that if we "spin" a point on the unit circle corresponding to the line with slope t, we reach the point corresponding to the line with slope -1/t. Set f(t) = -1/t; then f is a rational function of order 2 with the property that A(f(t)) = A(t). Likewise it follows from simple trigonometry that if we "swap" the point corresponding to a slope t = tan(theta) we arrive at the point corresponding to slope t' = tan(Pi/4 - theta) = (1-t)/(1+t). Set g(t) = (1-t)/(1+t); then g is a rational function of order 2 , commuting with f, with the property that A(g(t)) = A(t). Thus we now know that if A(t) = r for some t, there are three other solutions to this equation: t' = f(t), g(t), and g(f(t))=f(g(t)). Since it's clear that the equation A(X) = r may be unscramble to a quartic polynomial equation P(X) = 0 in X, these are then the only roots to P. If for some value of r this P(X) happens to be irreducible over Q, then we see that any extension field of Q containing one root t will contain all four of them: it's a splitting field for P. Moreover, the functions f and g generate a group of four permutations of the roots, which must then be the whole Galois group of P: Gal(P) = (Z/2Z) x (Z/2Z). (Actually while Galois groups over Q are probably the easiest to understand, it's more useful to us to note that what we have proved is that the field Q(t) of rational functions in t is Galois over the subfield Q( A(t) ) generated by this particular rational function, the Galois group again being G = (Z/2Z)x(Z/2Z). ) Now, let's push on further. The group G has three subgroups of order 2 (all normal, of course), so that if K/k is a Galois extension with this group, there should be three intermediate subfields, each quadratic over k. We can obtain them as the subfields K^ invariant under any of the three elements sigma of order 2 in G. Since these intermediate fields are just quadratic over k, we may write them as k' = k[ y ] where y is any element of K invariant under sigma but not in k itself. It's easy to create sigma-invariant elements; for example, for any x in K both x * sigma(x) and x + sigma(x) are sigma-invariant (since sigma is an automorphism of K of order 2.) For example, consider sigma = f (the function sending any of the four roots t to -1/t.) In this case t * f(t) = -1 is certainly f-invariant, but lies in the ground field Q. On the other hand, y = t + f(t) = (t-1/t) is also f-invariant, and [in case K = Q(t) is the function field, or in case k=Q and r is "generic"] not in the base field k. So we have a tower of fields K = Q(t) k' = Q( y ) where y = ( t-1/t ) k = Q( A(t) ) These should both be quadratic extensions. Clearly K is quadratic over k'; K = k'(t) where t satisfies the quadratic polynomial X^2 - y X - 1 in k'[X]. The other extension is just a little harder: we can show y is quadratic over k by showing it satisfies a quadratic polynomial over that field. Hmm, let's see, what would be the other root? Ah yes, the Galois group! The quotient group G/ should act here, and indeed it's generated by [the coset containing] g (="swap"). If we apply g to y we get g(y) = g(t) - 1/g(t) = (1-t)/(1+t) - (1+t)/(1-t) = -4t/(1-t^2). So the minimal polynomial of y over k should be (X-y)(X-g(y)) = X^2 - (t-1/t + (1-t)/(1+t)-(1+t)/(1-t)) X + (t-1/t)( -4t/(1-t^2) ) = X^2 - (-1/A(t)) X + 4 which is indeed in the ground field k = Q( A(t) ). If that went by a little too fast, let me observe that it just says we may solve A(t) = r by solving two quadratic equations in turn: first we solve X^2 + (1/r) X + 4 = 0 to get some root X = y; then we solve the equation t-1/t = y (i.e. t^2 - y t - 1 = 0) to get t itself. I only use Galois theory to do the grunt work for me of computing the two quadratic polynomials. Likewise we develop the intermediate field corresponding to sigma = g. This time we might for example select a g-invariant element to be z = t + g(t) (or t*g(t) would also work this time). Then k" = Q(z) should again be a field intermediate between K and k. The equation for K over k" is easily found to be t^2 -(z+1) t + 1 = 0. Again we find the minimal polynomial for z over k by computing z = (t^2+1)/(t+1) and so f(z) = (f(t)^2+1)/(f(t)+1) = (t^2+1)/t/(t-1); the polynomial is then (X-z)(X-f(z)) = X^2 - (-1/A(t)) X + (-1/A(t)), clearly in the ground field as needed. Finally I note that k(y,z) includes both intermediate fields, and hence has degree at least four over k, and thus must be K itself; in other words, we should be able to express t as a rational function of y and z. This isn't too hard; I get t = (z-2)/(y-z). Now, what has all this to do with rational right triangles? Well, I am interested in finding out when A(X) = r has rational solutions X=t. What my computations above have shown me is that on the one hand if there is a solution t, then there will be solutions y=t-1/t and z=(t^2+1)/(t+1) (thus both rational if t is) to the two quadratic equations X^2 + (1/r) X + 4 and X^2 + (1/r) X - 1/r. Conversely, if these two equations have rational solutions y and z respectively, then t=(z-2)/(y-z) is a rational solution to A(X) = r. So my single (fourth-degree) equation is equivalent to two quadratic equations. This isn't quite as lame a conclusion as you might think: it wouldn't have come out this way, for example, if the equation A(X)=r determined a Galois extension with group Z/4Z. Now, finally, I make the observation that, by using the quadratic formula, we see these quadratic equations have solutions iff certain elements of the underlying field are squares, namely the two discriminants "b^2-4ac". These are, respectively, equal to (1-4r)(1+4r)/r^2 and (1-4r)/r^2, so they are both rational squares iff both (1-4r) and (1+4r) are. This proves the theorem. **************** So I haven't actually succeeded with the original questions at all. We'd like to know whether A(t) = A(t') + A(t'') may be solved rationally, for example. This is a single equation in three variables. The only contribution of this long post is that rather than face a single equation of the form t^4 + F1(t', t'') t^3 + ... + F4(t',t'') = 0 it is sufficient to solve a _pair_ of equations which simply state G1(t',t'') is a square and G2(t',t'') is a square You can amuse your computer by letting it range over the possible t' and t'' hunting for squares, although of course it's faster to use some sophisticated machinery to limit your search to pairs which you already know make G1 a square; you can then hope to make G2 square as well. dave ============================================================================== From: trund@dovercliffs.com (Tom Rund) Subject: Re: Triangle Areas, Rational Squares, and Galois Theory. Date: Sat, 05 Jun 1999 Newsgroups: sci.math On 4 Jun 1999 rusin@shavano.math.niu.edu (Dave Rusin) wrote: >sestnick@ariel.com (Steve Estnick) asked: >>Can the areas of three rational right triangles >>on the same hypotenuse be in arithmetic progression? > >So Bergmann's question is at least as hard as Estnick's. Actually >Bergmann's question implies another challenge: can we even find >three rational right triangles with hypotenuse 1 whose areas sum >(i.e. A1+A2=A3) ? Speaking of other challenges, notice that if we just ask for rectangles with rational edges and a common but not necessarily rational diagonal, then it appears empirically that no four such rectangles exist with areas in arithmetic progression. Now, this calls to mind the fact that there are no four squares with areas in arithmetic progression. So, I'm going to boldly make a completely unsupported generalization of these two facts. First, notice that any rectangle with integer edge lengths a,b corresponds uniquely to a point on a plot of (a^2 - b^2) versus (a^2 + b^2), as shown below. (Warning: fixed pitch font graphic approaching!): a^2-b^2| / | / | / | / __________|/_____________ |\ a^2 + b^2 | \ | \ | \ | \ The right-hand region in between the two diagonal lines represents the rectangles with a >= b. Lines of constant area are hyperbolas, asymptotic to the diagonal lines. The squares are the rectangles with a=b, so those all lie along the horizontal axis. On the other hand, a set of rectangles with a common diagonal all lie along a single vertical line. So here's my reckless conjecture: There do not exist four points on any single line (of any slope) in this plane corresponding to rectangles with areas in arithmetic progression. ============================================================================== From: cmarshall@wildsurfs.com (Craig Marshall) Subject: Re: Triangle Areas, Rational Squares, and Galois Theory. Date: [missing] Newsgroups: sci.math On Sat, 05 Jun 1999 trund@dovercliffs.com (Tom Rund) wrote: >First, notice that any rectangle with integer edge lengths a,b >corresponds uniquely to a point on a plot of (a^2 - b^2) versus >(a^2 + b^2), as shown below. (Warning: fixed pitch font graphic >approaching!): > > a^2-b^2| / > | / > | / > | / > __________|/_____________ > |\ a^2 + b^2 > | \ > | \ > | \ > | \ > >The right-hand region in between the two diagonal lines represents >the rectangles with a >= b. Lines of constant area are hyperbolas, >asymptotic to the diagonal lines. >The squares are the rectangles with a=b, so those all lie along >the horizontal axis. On the other hand, a set of rectangles >with a common diagonal all lie along a single vertical line. >So here's my reckless conjecture: There do not exist four points >on any single line (of any slope) in this plane corresponding to >rectangles with areas in arithmetic progression. Let's flesh this out a little. The claim is that there do not exist constants A,B and non-zero integers aj, bj for j=1,2,3,4 (assume aj >= bj) such that (aj^2 - bj^2) = A(aj^2 + bj^2) + B j=1,2,3,4 (1) and such that the four products (aj bj), j=1,2,3,4 are in arithmetic progression. Obviously in the special case A=B=0 this requires aj=bj, and so the claim in this case is that the four squares aj^2 can't be in arithmetic progression, which is known to be true. This corresponds to four point on the horizontal line of the a^2 + b^2 axis. It's less convenient working in these coordinates for vertical lines, because then the coefficient A would be infinite. A more natural set of coordinates for this problem would be the diagonal lines shown on the above figure, because they represent the degenerate cases when b=0 or a=0, respectively, which are not proper rectangles. Using those axes as our coordinates, we see that they represent the a^2 and b^2 projections, and straight lines in these coordinates map to straight lines in the original coordinates. Specifically, if we denote a straight line in the a^2, b^2 coordinates as aj^2 = M bj^2 + K (2) then this corresponds to a straight line in the form of (1) with the coefficients M - 1 2K A = ----- B = ----- M + 1 M + 1 The advantage of the MK coordinates is that they are not singular for any proper aspect ratio. So, the original claim can be expressed by saying there are no constants M,K and positive integers aj,bj, j=1,2,3,4 (with aj >= bj) such that that (2) is satisfied for all j and such that the four products (aj bj), j=1,2,3,4 are in arithmetic progression. It might be possible to prove (or disprove) some special cases of this conjecture. For example, consider the horizontal lines in the original drawing. These have the form a^2 = b^2 + K for some constant K. We know that with K=0 this is just the well-known theorem on four squares in arithmetic progression. With K not equal to zero we have other sets of rectangles, such as the line K=105, which contains the lattice points (a,b) = (53,52), (19,16), (13,8), (11,4). Each of these points corresponds to a factorization of 105. Obviously the areas of these particular rectangles are not in arithmetic progression. Can we prove that four such areas can never be in arithmetic progression for any given K?