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Slot-wise Addition of Pythagorean Triples

Date: 07/17/2003 at 13:28:12
From: Chester
Subject: Slot-wise addition of Pythagorean triples

Is it possible to have a primitive Pythagorean triple, (a,b,c) such 
that a^2+b^2 = c^2, and two other Pythagorean triples, not necessarily 
primitive (x,y,z) and (p,q,r) with the property that a = x+p, b = y+q, 
and c = z+r?

It is very hard to find an example to verify that it is possible, and 
when dealing with arguments involving even and odd numbers, you see 
that there is not a direct contradiction, which leaves the question 
open.

The work mainly deals with arguments involved with even and odd 
numbers, because in a primitive PT, (a,b,c), only one of a,b is even, 
and the other is odd, leaving c odd.


Date: 07/17/2003 at 18:11:16
From: Doctor Rob
Subject: Re: Slot-wise addition of Pythagorean triples

Thanks for writing to Ask Dr. Math, Chester.

I think that what you ask is impossible. My argument begins like this:

   x^2 + y^2 = z^2,
   p^2 + q^2 = r^2,
   (x+p)^2 + (y+q)^2 = (z+r)^2.

Expand the last, and subtract from it the first two. Divide by 2, and 
the result is

   p*x + q*y = r*z.

Now we know that p, q, r, x, y, and z satisfy

   p = k*(u^2-v^2),
   q = k*(2*u*v),
   r = k*(u^2+v^2),
   u > v > 0, u + v odd, GCD(u,v) = 1, k >= 1,
   x = K*(s^2-t^2),
   y = K*(2*s*t),
   z = K*(s^2+t^2),
   s > t > 0, s + t odd, GCD(s,t) = 1, K >= 1.

(Alternatively, the formulas for x and y might be swapped, which
leads to a separate case.)

Substituting, that implies that

   k*K*[(u^2-v^2)*(s^2-t^2) + (2*u*v)*(2*s*t)] =
                                        k*K*(u^2+v^2)*(s^2+t^2),
   (u^2-v^2)*(s^2-t^2) + (2*u*v)*(2*s*t) = (u^2+v^2)*(s^2+t^2),
   s^2*u^2 - s^2*v^2 - t^2*u^2 + t^2*v^2 + 4*s*t*u*v =
                          s^2*u^2 + s^2*v^2 + t^2*u^2 + t^2*v^2,
   0 = 2*(s^2*v^2 + t^2*u^2 - 2*s*t*u*v),
   0 = 2*(s*v - t*u)^2,
   s*v = t*u.

Now notice that s divides t*u, but GCD(s,t) = 1, so s divides u.
Write u = d*s, and then v = d*t.  Since GCD(u,v) = 1, this implies
that d = 1, so s = u and t = v.  Now you can easily see that

   a = x + p = (k+K)*(u^2-v^2),
   b = y + q = (k+K)*(2*u*v),
   c = z + r = (k+K)*(u^2+v^2),

so GCD(a,b,c) = k + K >= 2, and so (a,b,c) is not a primitive triple.
This contradiction eliminates this case.

I leave the other case to you.  It is simpler, and can be eliminated
by parity considerations.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Number Theory

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