Is Fermat's Last Theorem Solvable for n=4 Using Pythagorean Triples?

[mente oscura]

Hello.

I share with you , a demonstration, which authorship is mine.

I do not know, if it exists, similar other one.Section A) demonstration Fermat’s Last Theorem, for n = 4.

[tex]Let \ A, \ B, \ C \ \in{\mathbb{N}} / A, \ C \ = \ odd; \ B \ = \ even \ / A^2+B^2=C^2[/tex]

(*)We consider, as possible:

[tex]A=x^2[/tex]

[tex]B=y^2[/tex]

[tex]C=z^2[/tex]

So:

[tex]x^4+y^4=z^4[/tex]

According to the "traditional" formulas, to obtain "Pythagorean Triplet":

[tex]A=M^2-N^2[/tex]

[tex]B=2MN[/tex]

[tex]C=M^2+N^2[/tex]

Properties:

1ª)

[tex]C+B=M^2+N^2+2MN=(M+N)^2[/tex]

[tex]C+B=z^2+y^2=(M+N)^2[/tex]. Por (*)

2ª)

[tex]C-B=M^2+N^2-2MN=(M-N)^2[/tex]

[tex]C-B=z^2-y^2=(M-N)^2[/tex]. Por (*)

Conclusion: we get two "Pythagorean triplet", in which two of its elements are equal.

[tex]z^2+y^2=(M+N)^2[/tex]

[tex]z^2-y^2=(M-N)^2[/tex]

Consideration: If we show that it is not possible that two "Pythagorean triplet", have two common elements, we will have shown the Fermat Last Theorem, for n = 4.

Section B) demonstration of the impossibility of that, two "Pythagorean triplet", have two common elements:

[tex]Let \ a, \ b, \ c \ \in{\mathbb{N}} / a, \ c \ = \ odd; \ b \ = \ even \ / a^2+b^2=c^2[/tex].(1)

[tex]Sean \ d, \ b, \ c \ \in{\mathbb{N}} / d, \ c \ = \ odd; \ b \ = \ even \ / c^2+b^2=d^2[/tex].(2)

Considerations:

1º) a, b, c, d, co-prime, so are primitive Pythagorean Triplets.

2º) “c” It is the odd small, which meets (1) y (2)Option 1º)

[tex]a^2+b^2=c^2[/tex]

[tex]a=m^2-n^2[/tex]

[tex]b=2mn[/tex]

[tex]c=m^2+n^2[/tex]Option 2º)

[tex]c^2+b^2=d^2[/tex]

[tex]c=u^2-v^2[/tex]

[tex]b=2uv[/tex]

[tex]d=u^2+v^2[/tex]

Course: "m" and "n" co-prime, same as "u" and "v".

By Option 1º) y Option 2º):

[tex]b=2mn=2uv[/tex]

I'm going to break down "b" in four co-prime factors: "e", "f", "g" and "h", such that:

[tex]b=2efgh[/tex]

And, we assign, for example:

[tex]m=ef[/tex]

[tex]n=gh[/tex]

[tex]u=eg[/tex]

[tex]v=fh[/tex]

Being: e=mcd(m,u), f=mcd(m,v), g=mcd(n,u), h=mcd(n,v)

Note: “mcd”= greatest common divisor.

Now, using the other element common to the Pythagorean Triplets: "c"

[tex]c=m^2+n^2=e^2f^2+g^2h^2[/tex]

[tex]c=u^2-v^2=e^2g^2-f^2h^2[/tex]

Therefore:

[tex]e^2f^2+g^2h^2= e^2g^2-f^2h^2[/tex]

[tex]g^2h^2+f^2h^2= e^2g^2-e^2f^2[/tex]

[tex]h^2(g^2+f^2)=e^2(g^2-f^2)[/tex]

To be "h" and "e" co-prime, and, also, [tex]g^2+f^2[/tex] y [tex]g^2-f^2[/tex], (note that "g" or "f" should be "evenr"), it follows that:

[tex]h^2=g^2-f^2[/tex]

[tex]e^2=g^2+f^2[/tex]

Thus, we have other two Pythagorean Triplets with two common elements, but:

[tex]c \ > \ g[/tex]

Since:

[tex]c=m^2+n^2>n^2=g^2h^2>g^2>g[/tex]

It is not possible, since we had considered "c", as the odd minor who met the conditions (1) and (2).

Whereupon, demonstrated the Section B)" and, therefore, also the "Section A)", concluding the impossibility of:

[tex]x^4+y^4=z^4[/tex]


Regards.

 

[mathbalarka]

I haven't checked through all of your proof, but yes, your approach is absolutely correct : the main step towards FLT4 is to prove that there can't be any 4-tuple $(a, b, c, d)$ satisfying $a^2 + b^2 = c^2$ and $b^2 + c^2 = d^2$. The standard step actually is to rearrange and multiply both sides to get $(ad)^2 = c^4 - b^4$ and it can be shown by infinite descent that $x^4 - y^4 - z^2$ over $\Bbb C^3$ has no integer points over it.

EDIT : Hold it. I don't understand your step when you break up $b$ in 4 coprime factors $e, f, g$ and $h$. $b = 2mn = 2uv$ is absolutely correct and upto that I agree, and as a conclusion $mn = uv$, but I am not sure why you are assuming that $(m, n, u, v) = (ef, gh, eg, fh)$. How about $(m, n, u, v) = (pqr, xyz, pxy, qrz)$? In that case $m$ and $n$ are coprime, so are $u$ and $v$ and $m \cdot n = u \cdot v = xyzpqr$ also.

So my point is why break $b$ in $\boxed{4}$ parts? Why not $6$ or $8$ parts, like I did above?

 

[mente oscura]

I haven't checked through all of your proof, but yes, your approach is absolutely correct : the main step towards FLT4 is to prove that there can't be any 4-tuple $(a, b, c, d)$ satisfying $a^2 + b^2 = c^2$ and $b^2 + c^2 = d^2$. The standard step actually is to rearrange and multiply both sides to get $(ad)^2 = c^4 - b^4$ and it can be shown by infinite descent that $x^4 - y^4 - z^2$ over $\Bbb C^3$ has no integer points over it.

EDIT : Hold it. I don't understand your step when you break up $b$ in 4 coprime factors $e, f, g$ and $h$. $b = 2mn = 2uv$ is absolutely correct and upto that I agree, and as a conclusion $mn = uv$, but I am not sure why you are assuming that $(m, n, u, v) = (ef, gh, eg, fh)$. How about $(m, n, u, v) = (pqr, xyz, pxy, qrz)$? In that case $m$ and $n$ are coprime, so are $u$ and $v$ and $m \cdot n = u \cdot v = xyzpqr$ also.

So my point is why break $b$ in $\boxed{4}$ parts? Why not $6$ or $8$ parts, like I did above?

Hello.

where is the problem?

[tex]b=2mn=2uv=2xyzpqr[/tex]

According to your example:

[tex]m=pqr[/tex]

[tex]n=xyz[/tex]

[tex]u=pxy[/tex]

[tex]v=qrz[/tex]

We operate:

[tex]c=m^2+n^2=p^2q^2r^2+x^2y^2z^2[/tex]

[tex]c=u^2-v^2=p^2x^2y^2-q^2r^2z^2[/tex]

[tex]p^2q^2r^2+x^2y^2z^2=p^2x^2y^2-q^2r^2z^2[/tex]

[tex]q^2r^2(p^2+z^2)=x^2y^2(p^2-z^2)[/tex]

[tex]q^2r^2=p^2-z^2[/tex]

[tex]x^2y^2=p^2+z^2[/tex]

Two Pythagorean Triplets with two common elements, but:

[tex]c \ > \ p[/tex]

[tex]c=m^2+n^2=p^2q^2r^2+x^2y^2z^2[/tex]

Therefore:

[tex]c \ > \ p^2 \ > \ p[/tex]

Regards.

 

[mathbalarka]

Excellent, you have made it clear that even $4$ or $8$ or higher factors would result the same. I'll move on to the next point :

[tex]e^2f^2+g^2h^2= e^2g^2-f^2h^2[/tex]

[tex]g^2h^2+f^2h^2= e^2g^2-e^2f^2[/tex]

[tex]h^2(g^2+f^2)=e^2(g^2-f^2)[/tex]

To be "h" and "e" co-prime, and, also, [tex]g^2+f^2[/tex] y [tex]g^2-f^2[/tex], (note that "g" or "f" should be "evenr"), it follows that:

[tex]h^2=g^2-f^2[/tex]

[tex]e^2=g^2+f^2[/tex]

I don't understand this bit. $h$ and $e$ are coprime, yes, as $(m, n) = 1$, but why are $g^2 + f^2$ and $g^2 - f^2$ relatively coprime too? Please clarify.

[Note : If $ab = xy$, it is necessary condition to have $(a, x) = (b, y) = 1$ to conclude $a = y$ and $b = x$. Take for example, $5 \cdot 4 = 2 \cdot 10$, where $4$ and $10$ have nontrivial gcd]

 

[mente oscura]

Excellent, you have made it clear that even $4$ or $8$ or higher factors would result the same. I'll move on to the next point :
I don't understand this bit. $h$ and $e$ are coprime, yes, as $(m, n) = 1$, but why are $g^2 + f^2$ and $g^2 - f^2$ relatively coprime too? Please clarify.

[Note : If $ab = xy$, it is necessary condition to have $(a, x) = (b, y) = 1$ to conclude $a = y$ and $b = x$. Take for example, $5 \cdot 4 = 2 \cdot 10$, where $4$ and $10$ have nontrivial gcd]

Hello.

I have considered that "g" or "f" is an even number. Only one.

You can choose, what is the pair, and see how the result is similar.

In this case, the show is trivial:

[tex]Let \ i \ , \ j \in{\mathbb{N}} \ / one \ is \ even \ and \ the \ other \ odd[/tex]

[tex]If \ k |(i+j) \ and \ k |(i-j) \rightarrow{} k | (i+j+i-j) \rightarrow{}k|i[/tex]

[tex]If \ k |(i+j) \ and \ k |(i-j) \rightarrow{} k | (i+j-i+j) \rightarrow{}k|j[/tex]

Conclusion:

"i" and "j" they are not coprime.

Regards.