A recent post at jd2718 noted that for all Pythagorean triples , that is to say, for all positive integer solutions to , it turns out that 60 divides the product .

In the course of exploring proofs of that result, I suggested a proof using the representation , which generates all primitive Pythagorean triples (triples which have no common factors). Proving that 60 divides their product proves the general result, as every Pythagorean triple has the form , where *k* is a positive integer and is primitive.

In the ensuing discussion, a question arose as to how we know that all possible Pythagorean triples can be found in this way. Since it is easier for me to post LaTeX code to this blog rather than in the comments on the jd2718 blog, I’ll present the proof here.

Suppose that , with positive integers having no common factors. This implies that exactly one of *a* and *b* is even (since if both are even, then so is *c*, leading to a contradiction, while if both *a* and *b* are odd, then both and are congruent to 1 mod 4, while would be congruent to 0 mod 4, not 2 mod 4).

So without loss of generality, assume that *a* is even. Thus is also even, and since and differ by an even number, *both* of those factors must be even.

Furthermore, exactly one of and is 2x(an odd number), since if both factors are divisible by 4, it would follow that both *b* and *c* would be even (since *b* and *c* are .) Likewise, any odd common factor of and would also then be a common factor of both *b* and *c*, from which it follows that

- and have no odd common factors
- one of these expressions is of the form (for some odd
*u*)
- and hence the other expression is of the form , where , where
*u* and *v* are positive integers

It now follows that , and . We can find *a* using , and thus .

Conclusion: every primitive Pythagorean triple has the form , for positive integers *u* and *v*.

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November 15, 2008 at 8:42 pm |

why must one of be of the form ?

I see twice an odd number, not twice an odd square. What have I missed?

(oh, and for the wordpress installation of LaTeX in wordpress comments, if it’s easy, I just do it. If it’s hard, I write a dummy post and preview it.. I still screw up, but not all the time)

Jonathan

November 16, 2008 at 9:35 am |

All of the prime factors of occur to even powers. Any odd prime factor of (say) cannot also be a factor of , and so must occur to an even power in . Thus both and must be a power of 2 multiplied by an odd square.

November 26, 2008 at 9:42 am |

I much prefer the classic geometric proof, based on stereographic projection. First, one notes that primitive Pythagorean triples are in bijective correspondence with rational pairs on the unit circle. So it suffices to parametrize rational points on the unit circle.

Consider the point where the line through and intersects the line . Since these two lines have rational coefficients, the point of intersection has rational coordinates, say [the factor of 2 makes the arithmetic come out nicer]. You can work out what is in terms of : just note that the point on the circle is the intersection of two lines, the first one through and , which is

and the second is the line perpendicular to that through , which is

.

[That is, we use here a famous result from Euclidean geometry: that a right triangle is formed when you draw the chords from a point on the circle to the endpoints of a diameter.]

Solving for the intersection of those two lines, one quickly finds

Putting , this gives

and voila! there’s your Pythagorean triple. Neat, isn’t it?

November 26, 2008 at 9:45 am |

Edit: the (x, y) in the fourth and fifth lines of the second paragraph should have been (u, v).

December 4, 2008 at 2:56 am |

[…] can learn how to bound binomial coefficients at the Endeavour, or generate Pythagorean triples at 360. To stretch your mathematical muscles a little more look for Terry Tao, considering polynomials […]

December 14, 2008 at 2:44 pm |

Here is the original Pythagorean formula: For any pair of positive integers “y > x” odd and coprime,

where the product “xy = i” is the odd leg and “(y2 − x2) / 2 = p” is the even leg,

being “(y2 + x2) / 2 = h” the hypotenuse and the radius of the incenter is “r = x(y − x) / 2”.

Remembering that “h − p = x2” is easy to get “h + p = y2” from a known primitive triple.

By the way “n + m = y”, then “(y − x) / 2 = n” and “y − ((y − x) / 2) = m”.

February 12, 2009 at 10:03 am |

(2 u v, u^2 – v^2, u^2 + v^2) (or (u v, (u^2 – v^2)/2, (u^2 + v^2)/2) for that matter) is not primitive for non-coprime u, v.

Harbey’s form is correct.

It is trivial to show that 8|(u^2 – v^2) for odd u, v

u^2 – v^2 = (u + v)(u – v)

=(2m + 1 + 2n + 1)(2m + 1 – 2n – 1)

=4(m + n + 1)(m – n)

2|(m + n + 1) or (m – n)

8|(u^2 – v^2)

4|((u^2 – v^2)/2)

The divisibility of one of the legs by 3 arises from the fact that 2 is not a quadratic residue mod 3. A similar approach shows that 5 divides one of the sides, whence the divisibility by 60 of the product of the sides.

February 24, 2009 at 6:12 pm |

[…] What would come next? Is there even an ordering for Pythagorean triples? (Well, maybe: TwoPi wrote earlier about how every Pythagorean triple can be written as , for positive integers u and v, so we could […]

March 7, 2009 at 5:43 pm |

Re:Juxtaposition: Mill’s question above: Is there an ordering for Pythagorean triples?

Please check out my hastily written online paper, “Ordering the Primitive Pythagorean Triples by Leg Difference and Size Using Generalized Pell Sequences”, in the Rational Argumentator!

March 7, 2009 at 6:33 pm |

Thanks Keith! I found a copy of your paper here.

January 23, 2010 at 9:02 am |

Here’s the revised and more formal version of that paper, published one year later in the Rational Argumentator:

http://rationalargumentator.com/issue232/PPTs_Pellian.html

April 17, 2009 at 11:33 am |

This question came up in a blog: Can there be two Pythagorean triples of the form {a,b,c} and {2a,d,c}. After much pondering and failed attempts to prove it impossible (see Natural Blogarithms — comments on Pythagorean triples — for details), my last response was this:

Fascinating question, Chris. It amounts to being able to stack two inscribed rectangles in a circle (of whole number radius) with the same heights and expect whole number lengths. Stacking on a horizontal diameter

We could generalize it to triples of the form {a,b,c}, {na,d,c} and get similarly inscribed rectangles, one with height n -1 times the other.

I suspect that not only would an irrational crop up, but that the big transcendental pi itself would come out to contradict this.

It also amounts to getting that the sinp = a/c, sinq =2a/c, cosp = b/c, cosq = d/c.

Anybody out there know if there’s a proof regarding this?

April 17, 2009 at 5:07 pm |

[…] speaking of reading things, after reading Keith Raskin’s comment I headed over to Natural Blogarithms to look for Pythagorean Triple stuff, and I was immediately […]

April 18, 2009 at 3:13 pm |

Answer (I think) to prime number problem referred to above by 360:

Time for a solution yet? Given A and B are primes.

Either A or B must be 2 in order for A + B to be prime (odd > 2)

It must be B in order for A – B to be prime (assuming primes are positive: 2, 3, 5 …)

The only prime bracketed between primes like this: A – 2, A, A + 2

is 5, since A must be 1 or 2 mod 3, and either A – 2 or A + 2 is 0 mod 3, divisible by 3, which is only okay if it is 3 itself. A can’t be 3, since 1 is not prime; that’s how I know it’s got to be 1 or 2 mod 3.

So, A = 5, b = 2

A + B + A – B + A + B = 3A + B = 17 prime.

May 14, 2009 at 8:47 am |

My latest thoughts: If there exist Pyth. triples of the form {a, b, c}, {2a, d, c}; and if we can assume (which I think we can) that they are primitive and that the a and 2a can be represented as the even legs in the Euclidean Algorithm, then

{a, b, c} = {n^2 – m^2, 2nm, n^2 + m^2}

{2a, d, c} = {v^2 – u^2, 4nm, n^2 + m^2}.

This implies that (n^2 + m^2)^2 – (4nm)^2 is a square, since it’s root must be v^2 – u^2 (an odd leg).

So, n^4 – 14n^2m^2 + m^4 is a square, while n^4 – 14n^2m^2 + 49m^4 must also be a square, as its a squared binomial.

This corresponds to x^2 – 14x + 1 and x^2 – 14x + 49 being squares, where x = n^2 and m= 1. They differ by 48. The only squares that differ by 48 are 1 and 49, 16 and 64, and 121 and 169, which respectively correspond to n^2 = 0, 15, and 20, none of which are permissible.

Don’t know if this really leads anywhere.

Anyone with knowledge or rumor of a proof, please come forward.

July 29, 2009 at 5:03 pm |

Pythagorean Triples Summary and Clarification:

Regarding the question about whether or not two Pythagorean triples of the form {a,b,c} and {2a,d,c} exist. I don’t think so — still don’t have a proof — but the question leads to lots of interesting questions about squares and rationals.

Let’s assume such triples exist. We can also assume without loss of generality that they are both primitive triples. If we now assume that a is odd, we get a contradiction. Observe:

{a,b,c} = {u^2 – v^2, 2uv, u^2 + v^2}

{2a,d,c} = {2xy, x^2 – y^2, x^2 + y^2}

for some u,v and x,y such that each pair is relatively prime and of opposite parity.

That means 2xy = 2a ⇒ xy = a, which means a is even, since either x or y is even.

Therefore a must be even and

{a,b,c} = {2uv, u^2 – v^2, 2uv, u^2 + v^2}

{2a,d,c} = {2xy, x^2 – y^2, x^2 + y^2}

where 2xy = 4uv and x^2 + y^2 = u^2 + v^2.

From here we get that c – 2a, c – a , c + a, and c + 2a are squares, because these are really

x^2 – 2xy + y^2, u^2 – 2uv + v^2, u^2 + 2uv + v^2, and x^2 + 2xy + y^2

OR (x – y)^2, (u – v)^2, (u + v)^2, and (x + y)^2.

I suspect that no set of squares are spaced out like that. In other words, I seriously doubt that there exist four squares of the form n^2, n^2 + k, n^2 + 3k, and n^2 + 4k. However, that’s just my mathematical intuition. I just don’t think that a parabola or a square root function will bend through those four linearly spaced points without hitting an irrational.

If such triples do exist, though, we would get the following bonus set of triples:

{4m, 4(u^2 – 4uv + v^2) – 1, 4(u^2 – 4uv + v^2) + 1}

{4(u – v), 4(u – v)^2 – 1, 4(u – v)^2 + 1}

{4(u + v), 4(u + v)^2 – 1, 4(u + v)^2 + 1}

{4w, 4(u^2 + 4uv + v^2) – 1, 4(u^2 + 4uv + v^2) + 1}

for m, w that equal the square roots of c – 2a, c + 2a, respectively.

August 30, 2009 at 10:53 pm |

LOOKS LIKE WE HAVE OUR PROOF (THANKS TO HENK)!

I buy Henk’s assumptions about the factors of c. They must be primitive hypotenuses as long as they are prime, which we can assume. This has been proven by Euler and the proof can be found in Richard Friedberg’s “An Adventurer’s Guide to Number Theory.”

AND it looks like the identities (below, also provided by Henk) provide our contradiction! If we assume that x and z are odd, then {xw – yz, yw + xz, kl} and {yw – xz, xw + yz, kl} represent

{a, b, c} and {d, 2a, c}. Therefore a = 2yz = xw – yz. So, 3yz = xw.

This is impossible, since x, y and w,z are relatively prime and x cannot be equal to w or z (nor can y be w or z). Yet if 3yz = xw, then x = 3z. But this forces y = w. Impossible!

Maybe I’m missing something, as I’m coming up with this as I’m writing this submission (which I really shouldn’t be doing!) But this looks like the contradiction I’ve been looking for.

THANKS HENK!!!

PROBLEM RESOLVED?

identities:

(x^2 + y^2)(w^2 + z^2) = (xw – yz)^2 + (yw + xz)^2

= (yw – xz)^2 + (xw + yz)^2

September 12, 2009 at 4:33 pm |

Re: My August 30th post

I think I thought I had good reason to assert that neither x nor y could equal to z or w, but i can’t see that reason at all now.

Don’t know if there’s a solid contradiction there after all.

February 25, 2011 at 3:12 pm |

HEY! How about a contest???

Who can either unearth or craft a proof that no pair of Pythagorean triples of the form {a, b, c} and {2a, d, c} exists?

Without loss of generality, I believe, we can safely assume that the triples are primitive and the doubled leg is the even one, ie a is even.

We could extend this to any or all n such that no pair of Pythagorean triples of the form {a, b, c} and {na, d, c} exists.

Whaddayou think???

July 1, 2011 at 9:17 am |

Full Disclosure: At the time I posted the above challenge a fairly esteemed blogger outlined a proof using the Fibonacci Box method that triples cannot be transformed to get the above result. I argued that I needed more than inability to transform; I needed proof of the impossibility of the existence of the triples above. We agreed to disagree.

But perhaps the blogger is right and there is a proof resting on the use of Fibonacci Boxes and/or Ternary Trees.

January 2, 2012 at 3:00 pm |

Please post the above comment under Keith Raskin.

Thanks!

January 2, 2012 at 3:41 pm |

The proof I’m discussing directly above using the Fibonacci Box method is definitely invalid for our purposes. We need to go beyond proving that modified triples will not result in the same hypotenuse, unless the modifications are totally unlimited, which they are not in the proof given.

June 17, 2012 at 11:05 am |

Corollaries and conjecture:

If a pair of triples of the form {a,b,c} and {2a,d,c} exist, then it is easy to show that

c – 2a, c -a, c + a and c + 2a are squares. Just look at their forms in Euclid’s algorithm,

given that a must be even (must equal 2mn, c – a = m^2 – 2mn + n^2,

c – 2a = r^2 – 2rs + s^2, etc.)

Let these squares be represented as M^2, N^2, O^2 and P^2. The first and last two

differ by a = N^2 – M^2, and the middle two differ by 2a. So, we can represent them

as M^2, N^2, 3N^2 – 2M^2 and 4N^2 – 3M^2, with M and N odd. It appears from

inspection of tables that when 3N^2 – 2M^2 or 4N^2 – 3M^2 is a square, the other

misses being an odd square by a factor of 8. In other words, either O or P must be irrational, it appears. In still other words, if you take an interval with perfect square

endpoints and advance it by double its length, one of the new endpoints is not a

perfect square; the image of the original interval under the square root function has

integral length, while the image of the new one has irrational length. If this conjecture is proven true, it would suffice in proving that the original triples could not

exist.

Also, if we represent O as N +q and P as N + p, then we get that q^2 + 2Nq over

p^2 + 2Np must equal 2/3. There are some inherent constraints: p and q are even

integers, with p bigger, and N must be greater than or equal to 1/2q^2 + Nq + 1.

If that quotient can never be exactly 2/3 for qualifying entries of p, q and N, then no

such original triples exist.

June 19, 2012 at 9:22 am |

Corollaries and conjecture:

If a pair of triples of the form {a,b,c} and {2a,d,c} exist, then it is easy to show that c – 2a, c -a, c + a and c + 2a are squares. Just look at their forms in Euclid’s algorithm, given that a must be even (must equal 2mn, c – a = m^2 – 2mn + n^2, c – 2a = r^2 – 2rs + s^2, etc.)

Let these squares be expressed as M^2, N^2, O^2 and P^2. The first and last two differ by a = N^2 – M^2, and the middle two differ by 2a. So, we can express them as M^2, N^2, 3N^2 – 2M^2 and 4N^2 – 3M^2, with M and N odd. It appears from inspection of tables that when 3N^2 – 2M^2 or 4N^2 – 3M^2 is a square, the other misses being an odd square by a factor of 8. In other words, either O or P must be irrational, it appears. In still other words, if you take an interval with perfect square endpoints and advance it by double its length, one of the new endpoints is not a perfect square; the image of the original interval under the square root function has integral length, while the image of the new one has irrational length. If this conjecture is proven true, it would suffice in proving that the original triples could not exist.

Also, if we represent O as N +q and P as N + p, then we get that q^2 + 2Nq over p^2 + 2Np must equal 2/3. There are some inherent constraints: p and q are even integers, with p bigger, and N must be greater than or equal to 1/2q^2 + Nq + 1. If that quotient can never be exactly 2/3 for qualifying entries of p, q and N, then no such original triples exist.