Natural Number Solutions of x^2 + y^2 = z^2


I always thought that the solution to Fermat's Last Theorem would , once found, be obvious and relatively simple. Well, it has been very difficult and has been worked on by many truly great mathmaticians without being solved until recently. The proof that was found did not fit my idea of what Fermat had in mind.
        Nine years ago during winter break I decided to give it a try. The following is what I had written. I have not seen this approach used before but I am not a math whiz so if anyone has seen it, please e-mail me.
        To summerize, what I did was to simplify x^2 + y^2 = z^2 to a set of two variable equations that defines all possible solutions in the set of natural numbers. My idea was that once that set was defined, then it could be proven that for x^n+y^n=z^n to to have solutions for n>2 they must be in the same set.
        OK here it is. What did I do wrong? E-mail me.

Proposition: There exist a set of equations of less than three variables that defines all natural number solutions to x^2 + y^2 = z^2 for x,y,z in the set of natural numbers.

Method: For any solution set of x^2 + y^2 = z^2 , x<z and y<z or, for some natural number d , z = x+d or z = y+d.

Take z = y+d and rewrite x^2 = z^2 - y^2 as a set of first degree equations derived below.

z^2 - y^2 = (z + y)(z - y)
z^2 - y^2 = (y+d + y)(y + d - y)
z^2 - y^2 = d*(2*y + d)
x^2 + y^2 = z^2 therefore x^2 = d*(2*y + d)

In order for x to be a natural number, d*(2*y+d) must be a perfect square. So only the d's and y's that make d*(2*y+d) a perfect square are the desired solutions. Since d can be any integer for example let d=1. By substitution, (2*y+1) must be a perfect square. Obviously all possible integral values of y will not make (2*y+1) a perfect square but some will. The smallest that will is 4 and the next is 12. The list is infinite. But what if d is not 1? You could go through the same steps and find y's that made d*(2*y+d) a perfect square but to simplify the process , if I let y = n*d for a natural number n, then I eliminate all of the y's that cannot make d*(2*y+d) a perfect square because of d not being a factor of y.

Then by substitution , x^2 = d*( 2*n*d + d) = d^2*(2*n + 1) and therefore (2*n + 1) must be a perfect square. The reason (2*n+1) must be a perfect square is this: If the prime factors of d are (p1,p2,…pn) the the prime factors of d^2 are (p1,p2,…pn)*(p1,p2,…pn) or (p1*p1,p2*p2…pn*pn) therefore ,even if (2*n+1) has some of the same prime factors as d, it must have products of primes also to make the product d^2*(2*n + 1) have paired products of prime factors. Therefore only permit the values of (2*n+1) that are perfect squares by leting 2*n + 1 = L^2 for some natural number L. (2*n + 1) is always odd; however by substitution x^2 = d^2 * L^2 and the positive solution for x is d*L so, if d is even, L may be even and still satisfy the same conditions. Therefore x=d*L for d and L both even or for any d with only odd L's.

Initially I let y = n*d and since 2*n+1 = L^2 then by sbstitution y = d(L^2 - 1)/2
Since z=y+d then by substitution z = d*(L^2 + 1)/2 + d or z = d*(L^2 + 1)/2
If you do not stipulate the odd or even conditions of L and d , then some solutions will be false, but all of the true solutions will be in the set of integers defined by :

x = L*d
y = d*(L^2 - 1)/2
z = d*(L^2 + 1)/2

Check : Does (L*d)^2 + [ d(L^2 - 1)/2]^2 = [d(L^2 + 1)/2]^2 ? It does. I will not show it here as it is a simple proof.

Since I have only excluded the values of x,y and z that will not permit x^2 = z^2 - y^2 to have integral solutions, I can state the following as a theorem.

THEOREM J : For the set of natural numbers n and L the only solutions of x^2 = z^2 - y^2 are contained in the set described by

x = L*d
y = d*(L^2 - 1)/2
z = d*(L^2 + 1)/2
Where d and L are natural numbers.

If you have a spread sheet, you can easily construct a table that cranks out a few x,y,z's that satisy x^2= z^2 - y^2 before your computer runs out of snuff. Hint : make "n" an absolute single cell and make "L" a column of odd integers or for only even n's make "L" a column of all integers. Also make a column of x^2 + y^2 next to the z^2column as a check that they are equal.

        If I haven't messed up somewhere , then Theorem J gives you some power to prove that if n is any even natural number greater that 2 then there are no solutions to x^n= z^n - y^n.

HYP H: There is no natural numbers x,y,z such that x^4 + y^4 = z^4 is true.
Proof: Assume that HYP H is false , then by Theorem J, the following three equations should define all possible integer solutions.

x = [L*d]^2
y = [d*(L^2 -1)/2]^2
z = [d*(L^2 + 1)/2]^2

If you make the sbstitutions and solve, then you get L=0 and L=1 as solutions. Therefore no non-trivial solutions are possible for x^4+ y^4 = z^4.

If I replace n in x^n= z^n- y^n with 2^n then the same proof would apply.
What about the case for n=6? By Theroem J

x = [L*d]^3
y = [d*(L^2 - 1)/2]^3
z = [d*(L^2 + 1)/2]^3

would have to define all possible solutions. Make the substitutions and see that there are no nontrivial solutions.


The End (for now)