wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> rational numbers
        
(Message started by: Ghost Sniper on Nov 7^th, 2007, 3:37pm)

Title: rational numbers
Post by Ghost Sniper on Nov 7^th, 2007, 3:37pm

Find such a rational number that when 5 is either added or subtracted from the square of the number, the result is the square of another rational number in both cases.

Title: Re: rational numbers
Post by towr on Nov 8^th, 2007, 2:39am

Finding one is simple enough, [hide]41/12[/hide], but I haven't been able to deduce it. So I also can't tell if there are any more yet..

Title: Re: rational numbers
Post by Joe Fendel on Nov 8^th, 2007, 2:10pm

Well, here is where I've gotten with this.

Let's imagine our number is (w/x) with w and x integers. Then (w/x)² + 5 is a square, as is (w/x)² - 5. Multiplying through by x² has w² + 5x² and w² - 5x² both squares, and since they're integers, they must be squares of integers, let's call them y² (for the larger one) and z² (for the smaller).

Then w² - z² = 5x², which can be factored into (w+z)(w-z). Let's set a=(w+z) and b=(w-z), so that w = (a+b)/2 and 5x² = ab. Then we have y² = ((a+b)/2)² + 5x² = (a² + 6ab + b²)/4, so that 2y=sqrt(a² + 6ab + b²).

Thus this "reduces" to finding a pair of integers a and b such that ab is 5 times a square and (a² + 6ab + b²) is four times a square. I say "reduces" because I'm not convinced this actually gets us any closer. But if we find such a pair, then set z=(a-b)/2, w=(a+b)/2, y=sqrt(a² + 6ab + b²)/2, and x = sqrt(ab/5), and we have (w/x)² + 5 = (y/x)² and (w/x)² - 5 = (z/x)².

Title: Re: rational numbers
Post by Grimbal on Nov 8^th, 2007, 3:03pm

A computer search for p/q with q<p<10000 reveals
[hide]41/12 as only solution in that range[/hide]

Title: Re: rational numbers
Post by Joe Fendel on Nov 9^th, 2007, 9:26am

The 12 isn't surprising. Going back to the equations in my earlier post, if we look at the fact that a²+6ab+b² = 4y² (mod 3), we get a² + b² = y², which means that either a or b is divisible by 3.

Also, if we look at this equation mod 16, suppose a and b are both odd. Then the left side is either 8 or 12 (mod 16) and the right side is 0 or 4 (mod 16). Thus either a or b is even, and then looking again at this equation (mod 2), we see that they are both even. However, at least one of them must be divisible by 4, otherwise we would have (a/2) and (b/2) both odd, and then the equation (a/2)² + 6(a/2)(b/2) + (b/2)² = y² would fail again (with the right side now equaling 0,1,4, or 9, but still not 8 or 12 (mod 16). Thus, ab is divisible by 24, and since ab = 5x², we know that x must be divisible by 12.

Title: Re: rational numbers
Post by Barukh on Nov 16^th, 2007, 3:55am

Excellent classical problem! I was surprised I didn’t find it on our forum. I don’t believe it belongs to Easy section, though.

Here’re some thoughts. Consider a Diophantine equation
x² – y² = 5z² (1)
We seek for 2 solutions (a, b, c) and (d, a, c), and then the required number is a/c. The idea is to find as many parametric solutions for the above quadratic as possible and check whether different solutions have two numbers in common, as required.

Assume x, y have the same parity (just a guess). Then, x+y = 2p, x-y = 2q, (1) becomes 4pq = 5z², and therefore z is even, say 2r, and pq = 5r² (following Joe’s steps).

Here, I tried to express p, q in terms of r. One possibility is to set p = 5r, q = r. This gives a family of solutions (3r, 2r, r). Another possibility is to set p = r², q = 5. This gives another family of solutions (r² + 5, r² – 5, 2r). In order for these solutions to form a needed pair, we should have one of the two: r² + 5 = 4r, or r² – 5 = 6r. Unfortunately, the first equation has no real roots, and the second has irrational roots. :(

What else could be done? Well, assume further that r = 2s. This yields pq = 20s². Set q = 4, p = 5s². This generates a third family (5s² + 4, 5s² – 4, 4s). Compare it with the second family, which can be written as: (4s² + 5, 4s² – 5, 4s). This immediately gives s = 3, and the needed triples are (a, b, c) = (41, 31, 12), (d, a, c) = (49, 41, 12). :)

I'm sure this can be done more elegantly.

Title: Re: rational numbers
Post by Barukh on Nov 19^th, 2007, 12:35am

I suggest one of the moderators moves this to at least Medium.

As I suspected, this problem is closely related to congruent numbers and (!) elliptic curves. The latter link suggests number of solutions is infinite.

Here's an example of another number satsifying the conditions: 3344161/1494696.

I will elaborate on how I found it later.

Title: Re: rational numbers
Post by Joe Fendel on Nov 19^th, 2007, 11:31am

:D Bravo, Barukh!

Title: Re: rational numbers
Post by Barukh on Nov 22^nd, 2007, 12:50am

on 11/19/07 at 11:31:18, Joe Fendel wrote:

:D Bravo, Barukh!

Thanks, Joe.

As promised, here’re some thoughts about how to derive more solutions from an existing one.

I will change notation in the following, since I will be considering rational numbers directly rather than integer numbers in my previous post. Also, I will consider the general case when the difference between rational squares is any rational number d.

So, we are seeking three rational numbers r, s, t, so that their squares form an arithmetic progression with common difference d. That is,

s² - r² = t² - s² = d (I)
Rearranging the terms and multiplying them, we can get the following relation: (rst)² = s⁶ - d²s². But then we have the following proposition:

If rational numbers r, s, t satisfy (I), then the point (s², rst) is a rational point on the elliptic curve¹ y² = x³ - d²x (II)

The good news is that there is a routine technique to generate rational points on EC if we know just one rational point with y different from 0. So we would like to have something converse to (II): given a rational point on the elliptic curve (EC), generate three numbers satisfying (I). Unfortunately, (II) can not be used directly for that.

But there is another trick to make this. Take Euclid’s characterization of Pythagorean triangles with free parameters x, d:
(x² - d²)² + 2xd² = (x² + d²),
and if (x, y) is a rational point at EC in (II),
2(2xd)(x² - d²) = 4y²d
Now adding (subtracting) the last two equations together, we easily get (try it!) (x² - d² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 2xd) ² = (x² + d²)² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 4y²d, and dividing both sides by (2y)²:
[(x² - d² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif 2xd) / 2y] ² = [(x² + d²) / 2y]² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif d
But that’s exactly what we need, and we have the following proposition:

If (x, y) is a rational point on EC in (II), then numbers r=(x²-d²-2xd)/2y, s=(x²+d²)/2y, t=(x²-d²+2xd)/2y satisfy (I) (III)
Let’s go back to our original question (d=5) and concentrate on propositions (II) and (III) only. We do have one triple satisfying (I), that is, r = 31/12, s = 41/12, t = 49/12. Using (II), we obtain a rational point (1681/144, 62279/1728) on EC (II) (check it!). Using this point, we get s = 3344161/1494696 from (III).

¹Elliptic Curves is an important object in modern math. Its usage ranges from FLT to modern cryptographic systems. In what follows, we are interested in finding rational points (http://www.asahi-net.or.jp/~KC2H-MSM/ec/eca1/) on such curves.