Is this result by coincidence or by implication?

[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.

 

[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.

 

[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.

 

[uart]

Good work dodo you've got it. :)

(Note that, in your example, c itself is a square, but this is not necessary for this proof.)

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

Substitute into ce = g^2 in order to prove ch = q^2

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

 

[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).

 

[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!