1999-2000
Pythagorean
Triples
Louis Nemzer
The Greek mathematician Pythagoras is best known for his formula that involves the sides of a right triangle: a2+b2=c2, where c is the hypotenuse of the right triangle. Part of Fermat’s last theorem involves the existence of whole number solutions to the Pythagorean theorem, the most familiar being {3,4,5}. The purpose of this paper is to find an algorithm that generates all of these solutions.
Fermat wrote his famous “last theorem” in the margin next to Diophantas’s method to solving this problem. However, this method does not address the issue regarding what will be defined as “prime triples.” This paper will suggest modifications to Diophantas’s method in order to resolve this.
Given:
a2+b2=c2
[Denoted "{a, b, c}" (where a<b<c)]
· A Prime Triple is a Pythagorean Triple in which the three terms are relatively prime.
· A Composite Triple is a Pythagorean Triple in which a positive integer divides all three terms (i.e. is a multiple of another triple)
· Any triple {aq, bq, cq} in which q is a positive integer is considered to be a multiple of {a, b, c}
· All triples must be prime or composite (See Divisibility Postulates)
Since there are an infinite number of multiples to every triple, the
first concern is to isolate only the set of prime triples. (This is why the
formula {3k, 4k, 5k} for
all whole numbers k can generate an infinite number of triples but be grossly
inadequate for our purposes as it deals only with one triple and its
multiples.)
Primary Postulate:
If d = (c-b), Then a2= d(2b+d)
a2 = c2 - b2
a2 = (c - b) (c + b)
a2 = d(c + b)
a2 = d(b + d + b)
a2= d(2b+d)
Prime Postulate:
If q is a positive integer and {a, b, c} is a triple, then {aq, bq, cq} is also a triple.
a2+b2=c2
a2q2 + b2q2 = c2q2
(aq)2 + (bq)2 = (cq)2
Divisibility Postulates:
If a prime q divides at least two of the terms of a triple, it must also divide the third. (Therefore, the triple is not prime)
If q is a prime integer, q/a, and q/b, then q/c
q/a2, q/b2
q/(a2 + b2)
q/c2
q/c
If q is a prime integer, q/a, and q/c, then q/b
q/a2, q/c2
q/(c2 - a2)
q/b2
q/b
If q is a prime integer, q/b, and q/c, then q/a
q/b2, q/c2
q/(c2 - b2)
q/a2
q/a
Even/Odd Postulate:
If d is defined as (c-b) where {a,b,c} is a prime triple, then in the prime factorization of d, all the even primes must have a odd power or zero, and the odd primes must have even powers.
Even Primes (2):
Suppose 2n/d, 2n+1/d,
& 2/n
a2= d(2b + d)
a2= 2d(b + .5d) [Recall, 2/d]
2d/a2 [2/a]
2n+1/a2
2n+2/a2
2n+2/2d(b + .5d)
2n+1/d(b + .5d)
2/(b+.5d)
2/b*
*Since 2/a and 2/b, by the divisibility postulates, the triple cannot be prime.
Odd Primes:
Suppose m is an odd prime, n is a
odd interger, mn/d, & mn+1/d.
a2= d(2b + d)
mn/a2 [m/a]
mn+1/a2
mn+1/d(2b + d)
m/(2b + d)
m/2b
m/b*
* Since m/a and m/b, by the divisibility postulates, the triple cannot be prime.
Colliery: In a prime triple, d must either be a perfect square or twice a perfect square
From the above, d must equal (xmynzo…) or (2kxmynzo…) where x,y, & z are odd primes; m, n, & o are even integers, and k is an odd integer.
Case
1:
d = (xmynzo…)
d = (xm/2yn/2zo/2…)2
Case 2:
d = (2kxmynzo…)
d = 2(2k-1xmynzo…)
d = 2(2(k-1)/2xm/2yn/2zo/2…)2
Diophantas's Method:
For any positive integers u and v:
{u2-v2, 2uv, u2+v2} is a triple.
It is simple to show that Diophantas’s method will only generate triples that had a d that is capable of generating prime triples (Either a perfect square or twice a perfect square).
If u2-v2>2uv
D= (u2+v2)-(u2-v2)
D= 2u2
Since u must be a whole number, 2u2 is twice a perfect square.
If u2-v2<2uv
D= (u2+v2)-2uv
D= (u-v)2
Since u and v are whole numbers and u>v, (u-v)2 is a perfect square.
However, this does not guarantee that a triple generated this way will be prime. For example:
Conjecture: A triple generated by diophantas's method is prime if and only if:
1. u and v are relatively prime
2. u+v is odd
1. Suppose q/u and q/v.
Q/u2, Q/v2
Q/u2-v2, Q/2uv, Q/u2+v2 *
2. Suppose 2/(u+v)
2 Cases: Either A. 2/u and 2/v OR
B. 2/u and 2/v
Case A. Suppose 2/u and 2/v
2/u2, 2/v2
2/u2-v2, 2/2uv, 2/u2+v2 *
Case B.
Suppose 2/u and 2/v
2/u2,
2/v2 [ODD2
= ODD]
2/(u2+v2) & 2/(u2-v2) [ODD+ODD=EVEN]
2/u2-v2, 2/2uv, 2/u2+v2 *