First we generate a formula for all Primitive Pythagorean Triples.
A Primitive Pythagorean Triple (PPT) is a triple of integers  p, q, r  for which
p 2  q 2  r 2 and the highest common factor of p, q and r is 1, i.e.
hcf  p, q, r   1
r
p
q
Since p 2  q 2  r 2 , any factor of two of these numbers is automatically a factor
of the third. Therefore, in any PPT, the highest common factor of any two of
the numbers is 1.
---------------------------It is easy to see, with a little algebra, that for any integers m, n the triple
m
2
 n 2 , 2mn, m 2  n 2  is a Pythagorean triple:
m
2
 n 2    2mn    m 2  n 2 
2
2
2
We now prove that every PPT can be written in this form.
Starting with p 2  q 2  r 2 , first notice that p and q cannot both be even, as
otherwise hcf  p, q, r   2 , and they cannot both be odd because
 2k  1   2l  1
2
2
is 2 more than a multiple of 4 and this cannot be equal to r 2
since all squares are of the form 4n or 4n  1 .
Therefore one of p and q is even and the other is odd (and r is odd).
Let’s say q is even. Then q can be replaced by 2u and p 2  q 2  r 2 can then
be written  2u   r 2  p 2 or, equivalently, 4u 2   r  p  r  p 
2
Since both r and p are odd then r  p and r  p are both even.
Let r  p  2 x and r  p  2 y . Substituting in 4u 2   r  p  r  p  gives u 2  xy .
The simultaneous equations r  p  2 x and r  p  2 y give r  y  x and
p  y x.
We have already established that hcf  r , p   1 . If x and y shared a common
factor greater than 1 then r  y  x and p  y  x would also share this
common factor. Consequently, hcf  r , p   1 implies that hcf  x, y   1 .
But the product of x and y is a square: u 2  xy . Since hcf  x, y   1 , both x
and y must themselves be squares since neither can pick up a matching
factor from the other.
With x  n 2 and y  m 2 we have:
q  2u  2 xy  2mn
p  y  x  m2  n2
r  y  x  m2  n2
Proof complete.
Finally, to generate all PPTs we need
 m  n (so that x  0 );
 hcf  m, n   1 (so that hcf  p, r   1 );

one of m and n is even and the other odd (to ensure p and r are both
odd) .
Back to the maths Item of the Month for June 2007
Let the two Pythagorean triples be
m
2
 n2 
m
2
 n2 
p
2
p
 q2 
2mn
2
 q2 
2 pq
Then the product of the largest values is
 m  n  p
and also  m  n  p
2
2
2
 q 2    mp  nq    np  mq 
2
2
2
 q 2    mq  np    mp  nq 
2
2
2
2
in the example given in the problem, we have m  2, n  1 to generate the triple
 3, 4,5 and p  3, q  2 to generate the triple  5,12,13 .
Using these values we have 5 13  82  12  7 2  42
Note that this is only guaranteed to work for primitive Pythagorean triples. For
example with the triples  9,12,15  which is not primitive and  5,12,13 you will
find that 13 15  195 cannot be written as the sum of two squares in even one
way never mind two.