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, = r4 - 2r²s² + s4 + 4r²s², = r4 + 2r²s² + s4, = (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((c-a)/(2d)).
There is a one-to-one 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.
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 (c-a)/2.
Let r = sqrt([c + a]/2), and
let s = sqrt([c - a]/2) are also integers, because...
(rs)² = r²s² = ([c+a]/2)([c-a]/2) = (c² - a²)/4 = (b/2)²
Since (b/2)² is a square and (c+a)/2 and (c-a)/2 are coprime (because the triple is primitive),
it follows that (c+a)/2 and (c-a)/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 vice-versa; 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 r²-1 = (r+1)(r-1) is a multiple of 3, and by the same logic so is s²-1. So the difference r²-s² is a multiple of 3, so a is a multiple of 3.
An interesting problem was posed on algebra-online.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.
The webmaster and author of this Math Help site is Graeme McRae.