
Suppose that we start with a primitive Pythagorean triple (a,b,c). If any two of a,b,c shared a common divisor d, then, using the equation
a² + b² = c²,
we could see that d² would have to divide the square of the remaining one, so
that d would have to divide all three. This would contradict the assumption that
our triple was primitive. Thus no two of
Now notice that not all of a, b, and c can be odd, because if
(2x+1)² + (2y+1)² = (2z+1)²,
4x² + 4x + 1 + 4y² + 4y + 1 = 4z² + 4z + 1,
4(x² + x + y² + y  z²  z + 1) = 3.
This implies that 4 is a divisor of 3, which is false. Now suppose that c
were even, and a and b odd, so
(2x+1)² + (2y+1)² = (2z)²,
4x² + 4x + 1 + 4y² + 4y + 1 = 4z²,
4(x² + x + y² + y  z² + 1) = 2.
This implies that 4 is a divisor of 2, which is also false. Thus either a or b must be even, and the other two odd. Let's say it is b that is even, with a and c odd. (If not, switch the meanings of a and b in what follows below.)
Now rewrite the equation in the form
b² = c²  a²,
(b/2)² = ([c  a]/2)([c + a]/2).
Notice that since b is even and a and c are odd, b/2,
This gives us the product of two whole numbers,
r² = (c + a)/2,
s² = (c  a)/2,
rs = b/2.
Furthermore, r > s, because
Now, solving the above three equations for a, b, and c, we find that, for primitive Pythagorean triples,
a = r²  s², b = 2rs, c = r² + s², r > s > 0 are whole numbers, r  s is odd, and the greatest common divisor of r and s is 1.
To see that the a, b, and c defined by these formulas do form a Pythagorean triple, just check the equation:
a² + b² = (r²  s²)² + (2rs)², = r^{4}  2r²s² + s^{4} + 4r²s², = r^{4} + 2r²s² + s^{4}, = (r² + s²)², = c².
The above formulas for a, b, and c are the most general formulas for
primitive Pythagorean triples. To every pair of whole numbers
r = sqrt([c + a]/2),
s = sqrt([c  a]/2).
Here is a table of the first few primitive Pythagorean triples:
r s a b c 2 1 3 4 5 3 2 5 12 13 4 1 15 8 17 4 3 7 24 25 5 2 21 20 29 5 4 9 40 41 6 1 35 12 37 6 5 11 60 61 7 2 45 28 53
To include all Pythagorean triples, both primitive and imprimitive, we let
a = (r²  s²)d, b = 2rsd, c = (r² + s²)d, r > s > 0 and d > 0 are whole numbers, r  s is odd, and the greatest common divisor of r and s is 1.
These formulas represent every Pythagorean triple. Given a Pythagorean Triple
(a,b,c) with common divisor d, and a/d and c/d are both odd, we can recover
d = GCD(a,c),
r = sqrt((c+a)/(2d)),
s = sqrt((ca)/(2d)).
There is a onetoone correspondence between Pythagorean triples and sets of
values of the three parameters
If a, b, c is a Primitive Pythagorean Triple such that a²+b²=c² then one of a,b is a multiple of three, and one of a,b is a multiple of four.
Proof:
We showed, above, that exactly one of a and b is even (if both are even, it's not primitive; if neither is even then a²+b² isn't a square) so without loss of generality, we assume b is even. That makes (c+a)/2 an integer, as well as (ca)/2.
Let r = sqrt([c + a]/2), and
let s = sqrt([c  a]/2) are also integers, because...
(rs)² = r²s² = ([c+a]/2)([ca]/2) = (c²  a²)/4 = (b/2)²
Since (b/2)² is a square and (c+a)/2 and (ca)/2 are coprime (because the triple is primitive),
it follows that (c+a)/2 and (ca)/2 are both squares. Also, it follows that r and s are coprime.
a = r²  s²,
and a is odd, so if r² is even, s² is odd, and viceversa; exactly one of r and
s is even.
b = 2rs, and since one of r and s is even, b is not only even, but it is a multiple of four as well.
That takes care of the "multiple of four" part of the proof. Now for the "multiple of three" part...
If r or s is a multiple of 3 then b is a multiple of 3. If neither one is a multiple of 3 then r²1 = (r+1)(r1) is a multiple of 3, and by the same logic so is s²1. So the difference r²s² is a multiple of 3, so a is a multiple of 3.
An interesting problem was posed on algebraonline.com in the beginning of September, 2000:
Prove that the area of a right triangle with integer sides is not a perfect square.
I didn't know how to do it at first, but I got some help. Here is the proof.
Prove that the area of a right triangle with integer sides is not a perfect square. I got some help, and finished this proof!
Puzzle question  if m is the product of n distinct primes, how many different right triangles with integer sides have a leg of length m?
Arithmetic Sequence of Perfect Squares, page 3  If a², b², c² are in arithmetic sequence, why is their constant difference a multiple of 24? Look at the second answer to this question for the relationship between Pythagorean Triples and this arithmetic sequence of squares.
Theorems Involving Perfect Squares  answers questions such as "why is the square root of x irrational unless x is a perfect square?" and other fundamental questions about perfect squares.
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