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Ghost Sniper
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rational numbers
« on: Nov 7th, 2007, 3:37pm » |
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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.
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towr
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Re: rational numbers
« Reply #1 on: Nov 8th, 2007, 2:39am » |
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Finding one is simple enough, 41/12, but I haven't been able to deduce it. So I also can't tell if there are any more yet..
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Wikipedia, Google, Mathworld, Integer sequence DB
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Joe Fendel
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Re: rational numbers
« Reply #2 on: Nov 8th, 2007, 2:10pm » |
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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.
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Grimbal
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Re: rational numbers
« Reply #3 on: Nov 8th, 2007, 3:03pm » |
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A computer search for p/q with q<p<10000 reveals
41/12 as only solution in that range
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| « Last Edit: Nov 19th, 2007, 12:35pm by Grimbal » |
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Joe Fendel
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Re: rational numbers
« Reply #4 on: Nov 9th, 2007, 9:26am » |
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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.
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Barukh
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Re: rational numbers
« Reply #5 on: Nov 16th, 2007, 3:55am » |
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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.
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| « Last Edit: Nov 16th, 2007, 3:56am by Barukh » |
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Barukh
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Re: rational numbers
« Reply #6 on: Nov 19th, 2007, 12:35am » |
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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.
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Joe Fendel
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Re: rational numbers
« Reply #7 on: Nov 19th, 2007, 11:31am » |
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 Bravo, Barukh!
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Barukh
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Re: rational numbers
« Reply #8 on: Nov 22nd, 2007, 12:50am » |
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on Nov 19th, 2007, 11:31am, Joe Fendel wrote:
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  2xd) 2 = (x2 + d2)2 4y2d, and dividing both sides by (2y)2:
[(x2 - d2 2xd) / 2y] 2 = [(x2 + d2) / 2y]2 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 on such curves.
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| « Last Edit: Nov 22nd, 2007, 1:00am by Barukh » |
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