Math Is Fun Forum

(a²+b², a²-b², 2ab) with positive integers a and b with a>b are all Pythagorean triples. 
If (x,y,z) is a Pythagorean triple then so is (kx,ky,kz) for integers k>1.  And if I recall
correctly all Pythagorean triples can be obtained from these formulas.

Most folks have trouble coming up with more than 2 or 3 Pythagorean triples.  I have
taught students to come up with many more by simply:
  1) Think of a positive integer >1.
  2)  Double it.  That's the 1st of the triple.
  3)  Square the number obtained in step 1)
  4)  Add one to that square.  That's the 2nd of the triple.
  5)  Subtract one from that square.  That's the 3rd of the triple.

Of course that's just the triple above with b set equal to 1 and a>1.

I don't know him either.

Hi!  This is Dan's proof of 4 years ago from your link.

Let a,b,c denote the sides of triangle ∆ABC and P and A its perimeter and area respectively.
Note that Heron’s formula states that the area of ∆ABC is: A = √[s(s-a)(s-b)(s-c)]
Where s denotes the semi-perimeter, s = (a+b+c)/2.
To find all such triangles such that P=A we must have
√(a+b+c)(b+c-a)(a+c-b)(a+b-c) = 4(a+b+c).
Put x=b+c-a, y=a+c-b and z=a+b-c, where x,y,z are positive integers (true by the triangle
inequality). We see that Heron reduces to the Diophantine equation 16(x+y+z)=xyz.

We see from above that x,y,z must all be even integers. So, we may put x=2m, y=2n, z=2k to get,

  mnk=4(m+n+k)
  ⇒ mnk-4m = 4(n+k)
  ⇒ m = 4(n+k)/(nk-4)

Without loss of generality assume m≥n≥k, then we have 2m≥2n≥n+k ⇒ nk≤12.
Since 4<nk≤12 we may test integral values of n and k to find all such triangles.
Finally, we see that there is only 5 such triangles:
(a,b,c) = (5,12,13), (6,8,10), (6,25,29), (7,15,20), (9,10,17)
..................................................................................................................................

I followed pretty easily to the  three little lines starting with mnk=4(m+n+k) but had a little
difficulty figuring out why 4<nk≤12.  So I went back to the line mnk=4(m+n+k) and got it
from there.  First rewrite mnk=4(m+n+k) as  nk = 4(m+n+k)/m. 

Since m≥n≥k then 3m = (m+m+m) ≥ (m+n+k).  Therefore
                              4(3m)          4(m+m+m)          4(m+n+k)
            12 = 3*4 = _____    =     __________    ≥   ________   =  nk  because m≥n≥k.
                             
                                 m                    m                        m
implies  (m+m+m) ≥ (m+n+k).

Also rewriting mnk=4(m+n+k) as  4 = nk*[m/(m+n+k)]  we see that nk is being multiplied
by the quantity [m/(m+n+k)] which is less than one.  Hence nk>4.

So now we have the 4<nk≤12.

Testing all n≥k being bigger than four but less than or equal to 12 gives us the combinations
(n,k) ∈ {(12,1),(6,2),(4,3),(3,2),(4,2),(5,2),(3,3),(6,1),(8,1),(5,1),(7,1), (9,1),(10,1),(11,1)}
to test (14 of them).

Testing means substitute the n and k into mnk=4(m+n+k) and solve for m.  If the result is a
positive integer then double the m, n, and k to get the x, y, and z.  Then put these values
into the equations x=-a+b+c
                           y= a-b+c
                           z= a+b-c    and solve this system for a, b and c (the sides of the triangles).

Example:  n=4 and k=3 gives  m*4*3=4(m+4+3); that is, 12m=4m+28  so  m=28/8 = 7/2
                so this case doesn't give an integral value of m.  Case fails.

Example:  n=6 and k=1 gives  m*6*1=4(m+6+1); that is, 6m=4m+28  so  m=14
                This case gives an integral value for m so x=2m=28, y=2n=12 and z=2k=2.

Substituting these in to the three equations involving a, b, and c we obtain

                           28 = -a + b + c
                           12 =  a -  b  + c
                            2  =  a + b - c

Solving this system by addition/subtraction we obtain a=7, b=15, c=20  which is one of the
triangles that works.  (Adding each pair of equations eliminates two variables leaving the
third which is easy to solve.)

A nice solution to the problem, but Dan leaves out a few steps that one must scratch the
head about to fill in the gaps.     gotta

Hi Agnishom!

Boy! Did I misread the problem or what?  Sure was easy for the square, but a horse of a different
color for the triangles.  I think you will find that the only two Pythagorean triples that will work are
the first two that bobbym  listed.  The basic Pythagorean triples are of the form

                                           (u²+v², u²-v², 2uv) with 0<v<u.

(6,8,10) is a scalar multiple of (3,4,5); that is, (2*3,2*4,2*5) where u=2 and v=1.
(5,12,13) is a basic Pythagorean triple where u=3 and v=2.

I would have responded sooner, but my internet connection has been out.

Have a great day!

Answer:

Hi phrontister!

I had never noticed those symbols!  Thanks for pointing them out. 

Here's a thought:

In general the difference of two squares can be written as

                                    2    2       2            2    2             2
                             (x+i)  - x  =  x + 2xi + i  - x   =  2xi+i   =  i(2x+i).

Setting this equal to 7, the only solution for i and x both greater than zero is x=3 and i=1.

                  2   2      2     2
So this (x+i) - x   = 4  - 3   = 16-9 = 7.

So the (2k)^2 = 16 making k=2 or -2.
And the (2n+19)^2 must equal 9 so 2n+19=3 or -3
Hence 2n+19 = 3  or  2n+19=-3
              2n = -16  or     2n = -22
                n = -8   or       n = -11

Yep!  Privy to a bit of number theory, eh?  Would make Fermat smile in his grave.

How about a base (other than two) for which one can write any positive number in terms of using no coefficients greater than one?

Three digit numbers with two 3's and one 1 or with two 1's and one 3.  Only one of them is NOT a
prime.

13 --- a baker's dozen.

How about the largest number less than 100 that can only be written one way as a difference of
squares?

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