Is this result by coincidence or by implication?

1. Nov 8, 2008

uart

The other day someone gave me a puzzle. It was a geometry type puzzle but I was a able to reduce it to a set of pythagorean triads. Essentually the problem turned out to be equilvalent to the following.

Find natural numbers a, b and c (each less than 42) such that {a, b, c} forms a Pythagorean traid and {(c-a), b} are also the first two elements of a second Pythagorean triad.

That is, a^2 + b^2 = c^2 and (c-a)^2 + b^2 = n^2 for some natural number n.

Since it was given that the problem had a solution it was a pretty easy matter of testing a few PT's to find the one that worked and I quickly found {a, b, c} = {7, 24, 25}.

Later it caught my attention that, {(c+a),b . }, also formed a Pythagorean triad, though this was not a requirement of the original problem. At first I suspected that this might be an implication of the other conditions but I was unable to deduce any such implication. So I thought I'd post it here and see if anybody else knew of any reason for that last PT other than just coincidence.

2. Nov 8, 2008

uart

BTW. Here is the original puzzle if anyone is interested.

I cut a rectangle from a piece of A3 paper (approx 42 by 30 cm). Both the length and height of the rectangle are a whole number of cm and when this rectangle is folded such that the diagonally opposite corners align then it forms a pentagon in which all five sides are also a whole number of centimeters. Find the size of rectangle that I cut.

The answer is (c+a) by b, for the a, b and c given previously.

3. Nov 9, 2008

dodo

I have a proof but, since it's rather long, I'll just outline the path and let you complete it. (If somebody sees a simpler way, please post it!)

Suppose you are given that {a,b,c} and {c-a,b,d} are Pythagorean triples. From the second, expanding (c-a)^2 + b^2 = d^2 and substituting a^2 + b^2 by c^2, you will get at 2c(c-a) = d^2.

Now prove the following lemma: "if {x,y,z} is a Pythagorean triple and z is even, then x and y must also be even".

Let {e,f,g} be the Pythagorean triple {(c-a)/2, b/2, d/2}, and arrive to ce = g^2.

Repeat everything for the triple (yet to be proven Pythagorean) {c+a,b,q}: either it is not Pythagorean, or ch = q^2, where h is the integer (c+a)/2.

Now prove that, if ce is a square and ch is also a square, then e = h . k^2 for some integer k. (Note that, in your example, c itself is a square, but this is not necessary for this proof.) Substitute into ce = g^2 in order to prove ch = q^2.

4. Nov 9, 2008

uart

Good work dodo you've got it. :)

Yes I was wondering if that was part of the "co-incidence" or not.

ok that was the bit I was missing but I've got it now, many thanks.

5. Nov 9, 2008

uart

I think this is about as simple a proof as I can make of this.

Given
$$a^2 + b^2 = c^2$$
and
$$(c-a)^2 + b^2 = n^2$$

Expand the second equation and substitute for b^2 from the first to get,
$$2 c (c-a) = n^2$$

Consider the triple, {(c+a), b, $\sqrt{(c+a)^2 + b^2}$}

Clearly this is a Pythagorean triple if $(c+a)^2+b^2$ is a square number.

As before expand and substitute for b^2 to get,
$$(c+a)^2 + b^2 = 2 c (c+a)$$

$$\,\,\,\,\,\,\, = \frac{2 c (c^2 - a^2)}{c - a}\,\,\,\,\,\,\,\,$$ : difference of two squares.

$$\,\,\,\,\,\,\, = \frac{2 c b^2}{c - a}\,\,\,\,\,\,\,\,$$ : since b^2 = c^2 - a^2

$$\,\,\,\,\,\,\, = \frac{4 c^2 b^2}{n^2}\,\,\,\,\,\,\,\,$$ : substituting for (c-a) from the 3rd equation.

So $\sqrt{(c+a)^2 + b^2 }$ is rational and therefore integer (by the rational root theorem).

Last edited: Nov 9, 2008
6. Nov 10, 2008

dodo

$$\frac{4 c^2 b^2}{n^2}$$ is a square 'by eye'... it's the square of 2bc/n. And it can be verified (with some more lengthy expanding) that { (c+a), b, 2bc/n }, with $$n = \sqrt { (c-a)^2 + b^2 }$$, is a Pythagorean triple.

Anyway, that's certainly shorter than what I did!