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Topic: Special Pythagorean Triangle (Read 659 times) |
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Barukh
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Special Pythagorean Triangle
« on: Jul 27th, 2006, 6:31am » |
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Among all Pythagorean triangles with integer sides and area that is a square of an integer, what is the triangle with the smallest hypothenuse?
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| « Last Edit: Jul 27th, 2006, 6:32am by Barukh » |
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JohanC
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Re: Special Pythagorean Triangle
« Reply #1 on: Jul 27th, 2006, 2:04pm » |
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on Jul 27th, 2006, 6:31am, Barukh wrote:| Among all Pythagorean triangles with integer sides and area that is a square of an integer, what is the triangle with the smallest hypothenuse? |
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Yes, a very interestingly formulated question ...
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Icarus
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Re: Special Pythagorean Triangle
« Reply #2 on: Jul 27th, 2006, 2:27pm » |
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Clearly, if you have such a triangle and the legs have a common factor, you can divide it out to find another triangle with the same properties. So we can assume the triangle has relatively prime legs. It is well known that all relatively prime pythagorean triples (x, y, z) are of the form (r2 - s2, 2rs, r2 + s2), where r and s are relatively prime, and one or the other is even. The area of the triangle is xy/2 = (r+s)(r-s)rs. Since r and s are relatively prime (and one is even), the four values r+s, r-s, r, s must be pairwise relatively prime. Hence if the area is a perfect square, so must be all four values:
r = a2
s = b2
r-s = c2
r+s = d2
This requires 2 related pythagorean triples:
a2 = b2 + c2
d2 = a2 + b2 = 2b2 + c2.
We note that a, b, c, and d are relatively prime. By the result for pythagorean triples mentioned above, a and d are odd, and therefore so is c, while b is even. If we apply the P.T. result to the new triples, we discover that for some t,u,v,w we have
t2+u2 = a = w2 - v2, and 2tu = b = 2vw
So we need a solution to the paired equations: t2 + u2 + v2 = w2 and tu = vw with t and u relatively prime (and one even), and with v and w relatively prime.
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And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Sjoerd Job Postmus
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Re: Special Pythagorean Triangle
« Reply #3 on: Jul 28th, 2006, 9:03am » |
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Ok... we want the smallest hypothenuse.
triangle: a,b,c where c = sqrt(a^2 + b^2)
if ( gcd(a,b) > 1 )
{
// By dividing a and b by gcd(a,b), we find a smaller triangle  , which could possibly also satisfy the equation...
}
In other words, we're talking about a primitive pyth. triangle. (3,4,5) is, (6,8,10) isn't.
We're looking to minimise c (or c^2, as you wish), under the constraint that .5 * a * b = square of a number.
Now, this also means something different. What does it mean? It means that of every factor of (.5*a*b), we have a multiplicity of 0, 2, 4, 6, 8... Now, if a would have an odd multipl. of a factor, so would b, so we don't have a primitive triangle.
So, a is a square, while b would be twice a square. (or the other way around).
This makes the searching a lot easier ..
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Barukh
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Re: Special Pythagorean Triangle
« Reply #4 on: Jul 29th, 2006, 6:32am » |
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on Jul 27th, 2006, 2:04pm, JohanC wrote:| Yes, a very interestingly formulated question ... |
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JohanC
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Re: Special Pythagorean Triangle
« Reply #5 on: Jul 29th, 2006, 9:19am » |
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on Jul 27th, 2006, 2:04pm, JohanC wrote:
Yes, a very interestingly formulated question ... |
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on Jul 29th, 2006, 6:32am, Barukh wrote:
I mean, on the one hand it's a trick question, impossibly to solve as stated.
On the other hand, it gives a clue that wouldn't be there if it were formulated straightforward.
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Grimbal
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Re: Special Pythagorean Triangle
« Reply #6 on: Jul 29th, 2006, 10:59am » |
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on Jul 28th, 2006, 9:03am, Sjoerd Job Postmus wrote:
So, a is a square, while b would be twice a square. (or the other way around).
This makes the searching a lot easier .. |
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So we are looking for a=x2 and b=2y2.
a2 + b2 = c2 becomes x4 + 4y4 = c2
where x, y, c are integers, x, c are odd.
Let's see mod 16.
For x = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
x2 = 0 1 4 9 0 9 4 1 0 1 4 9 0 9 4 1 0
x4 = 0 1 0 1 0 1 ...
4x4 = 0 4 0 4 0 4 ...
x being odd, x4 = 1 (mod 16)
since c2 can not be 5 (mod 16),
4y4 must be 0 (mod 16), so y is even.
So we are looking for a = x2 and b = 4y2.
where x, y, c are integers, x, c are odd.
a2 + b2 = c2 becomes x4 + 16y4 = c2
Let's see mod 5.
For x = 0 1 2 3 4 5 ...
x2 = 0 1 4 4 1 0 ...
x4 = 0 1 1 1 1 0 ...
16x4 = 0 1 1 1 1 0 ...
If neither x or y are multiples of 5, c2 would be 1+1=2 (mod 5), but it can only be 0, 1, 4.
They can not be both a multiple of 5, so exactly one of x or y is a multiple of 5.
This could make the searching a bit easier .
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| « Last Edit: Jul 29th, 2006, 11:30am by Grimbal » |
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Sjoerd Job Postmus
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Re: Special Pythagorean Triangle
« Reply #7 on: Jul 30th, 2006, 12:57am » |
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Let's also see mod3.
For x = 0 1 2 3 ...
x2 = 0 1 1 0 ...
x4 = 0 1 1 0 ...
16x4 = 0 1 1 0 ...
At least ONE of x or y is a multiple of 3. But, they can not be both a multiple of three, (primitivity)... But, I can't tell which one must be.
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JohanC
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Re: Special Pythagorean Triangle
« Reply #8 on: Aug 7th, 2006, 2:03am » |
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on Jul 29th, 2006, 9:19am, JohanC wrote:I mean, on the one hand it's a trick question, impossibly to solve as stated.
On the other hand, it gives a clue that wouldn't be there if it were formulated straightforward. |
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Hi, Barukh,
Probably you have a nice proof of the impossibility of such a triangle. All I can come up with is a quite large proof, not so suited for the medium forum.
Does your proof fit into 50 lines?
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Barukh
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Re: Special Pythagorean Triangle
« Reply #9 on: Aug 18th, 2006, 9:11am » |
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Quote:Hi, Barukh,
Probably you have a nice proof of the impossibility of such a triangle. All I can come up with is a quite large proof, not so suited for the medium forum.
Does your proof fit into 50 lines? |
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The proof I know begins by showing that if such a triangle existed, then we would have 2 squares whose sum and difference are also squares (trivially). The latter is shown to be impossible by the method of infinite descent (no too easy).
Is your proof similiar?
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Eigenray
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Re: Special Pythagorean Triangle
« Reply #10 on: Aug 18th, 2006, 3:41pm » |
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on Jul 27th, 2006, 2:27pm, Icarus wrote:a2 = b2 + c2
d2 = a2 + b2 = 2b2 + c2.
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This leads to
a4 - b4 = (a2-b2)(a2+b2) = c2d2.
Since Barukh mentioned infinite descent, we may proceed similar to Fermat proving x4+y4=z2 impossible:
| hidden: | First note that in Icarus's post, b,c relatively prime forces a odd, forcing b even, forcing z=cd odd. We now show there are no (non-zero integer) solutions to
a4 - b4 = z2 (*)
with z odd (in fact there are none with z even either). Suppose (a,b,z) is a solution to (*), with z odd. We find another solution with smaller |z|. As usual, we may assume a,b,z are pairwise relatively prime: if a prime p divides any two, by (*) it divides the third, and if p|a,b, then p4|z2, hence p2|z, so (a/p, b/p, z/p2) is a smaller integer solution.
First, since the RHS of (*) is 1 mod 4, we must have a odd, and b even. Then the primitive Pythagorean triple
(z,b2,a2) = (p2-q2, 2pq, p2+q2),
where, replacing z by -z if necessary, we may assume p odd, q even. Since 2pq = b2, and p,q are relatively prime, then in fact
p=r2, q=2s2.
Now, in the primitive Pythagorean triple
(p,q,a) = (m2-n2, 2mn, m2+n2),
q=2s2 = 2mn, so that
m = u2, n=v2,
and then
r2 = p = m2-n2 = u4 - v4,
meaning (u,v,r) is a solution to (*). Now, r is also odd, and
z = p2-q2 = (p+q)(p-q) > p+q > p = r2 > r.
Reductio ad absurdum.
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Aha. I knew this problem seemed familiar. Curiously, I adapted the same theorem in two different ways.
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| « Last Edit: Aug 18th, 2006, 4:01pm by Eigenray » |
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