1. No category

Beal’s Conjecture, Elegant Proof
Carlos Giraldo Ospina
December 21, 2014
Abstract
In this paper we prove that the equations of Beal’s conjecture
(BC) and Fermat’s last theorem (FLT) are equivalent. This implies
that proving FLT means proving BC and vice versa. We give a general
proof of FLT for exponents that are multiples of 4. Such proof becomes
a general proof for any exponent greater than 2. We also provide a
general proof for odd exponents. Each proof is less than one page
long, and so simple that any amateur or professional mathematician
is able to tell whether it is correct or incorrect.
The first step consists in using the well-known formula for
generating Pythagorean triples as well as reductio ad absurdum for
exponents that are multiples of 4, and we also use an equivalence
relation for any even or odd exponent. The method of reductio ad
absurdum is also applied to odd exponents.
Keywords: Beal’s conjecture, Fermat’s last theorem, Pythagorean triples
2010 Mathematics Subject Classification: 00-XX · 00A05 · 11-XX ·
11Dxx
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Introduction
Some authors state that FLT was proved by Fermat by using the method
of infinite descent for the exponent 4, while others state that such proof has
nowhere been found. Some people say that one of Fermat’s fans obtained
that proof and attributed it to Fermat.
Andrew Wiles stated: “I don’t believe Fermat had a proof. I think he
fooled himself into thinking he had a proof. But what has made this problem
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special for amateurs is that there’s a tiny possibility that there does exist an
elegant 17th-century proof.” (NOVA ONLINE)
Wiles states that there is “tiny possibility” that there exists an “elegant
proof”. Surprisingly, we now realize that professional mathematicians have
missed something that is within the reach of any average high school student.
But neither Fermat, nor Wiles, nor any another mathematician devoted to
FLT have noted that the general proof for multiples of 4 can be given in
less than eight lines. In the end of the proof we get to the absurd statement
“even equals odd”.
Both amateur and professional mathematicians will now have a proof of
FLT for exponents that are multiples of 4 as well as another proof for odd
exponents. These proofs can even be understood by average high school
students.
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Preliminaries
A. Equivalence between the equations of Beal’s conjecture and FLT
(xp + y q = z r ) ≡ [xp + (y q/p )p = (z r/p )p ]
≡ [(xp/q )q + y q = (z r/q )q ]
≡ [(xp/r )r + (y q/r )r = z r ]
In the first equivalence we can set y q/p = u and z r/p = v, and we get
(xp + y q = z r ) ≡ (xp + up = v p ).
In the second equivalence we can set xp/q = h and z r/q = k, and we get
(xp + y q = z r ) ≡ (hq + y q = k q ).
In the third equivalence we can set xp/r = e and y q/r = d, and we get
(xp + y q = z r ) ≡ (er + dr = z r ).
Since the equations of Beal’s conjecture and FLT are equivalent, it follows
that Beal’s conjecture is equivalent to FLT. In other words, proving FLT
means proving Beal’s conjecture and vice versa.
But we need to prove FLT by using a method that is different to that
employed by Dr. Wiles if we do not want to share the Beal Prize with that
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famous mathematician. Wiles should probably be given almost the whole
amount of the prize. Why is that? One thing is the equivalence between the
equations and another thing is the equivalence between the conjectures and,
as a result, we must prove one or the other.
B. Pythagorean equation
The Pythagorean equation is a basic tool we will use in order to give an
elementary proof of FLT. For this reason, it is necessary to analyze some
aspects of the mentioned equation.
We will not make use of any triangle (there are those who always associate
the Pythagorean equation with the right triangle). What the Pythagorean
theorem states is that in any right triangle the square of the length of the
hypotenuse is equal to the sum of the squares of the lengths of the other two
sides, which is not the same as saying that the Pythagorean equation has its
origin in the right triangle.
To sum up, FLT is just a matter of arithmetic, algebra, and numbers; it
is not a matter of physical objects. You can write the simplest or the most
complex equation you can think of, and if there is any physical object that
can be associated with your equation... congratulations!
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Generator of Pythagorean equations (GPE)
If x = 2s (where s is arbitrary), y = s2 − 1, and z = s2 + 1, then z 2 = x2 + y 2 .
The previously mentioned system will be referred to as “GPE”. By using
this system we obtain lots of generators of Pythagorean triples. On the other
hand, if we start with any generator we can “work backwards from there”
and get to the GPE again.
The well-known system
x = 2tr, y = t2 − r2 , z = t2 + r2 ⇒ z 2 = x2 + y 2
(1)
can be deduced from GPE if we set s = t/r and we eliminate denominators
(this is different from the concept of primitive Pythagorean triples):
t2
+1
r2
2
2 2
2
2
2 2 2
2
t
t
t
t
t
= 2
+
−1 ⇒
+1 = 2
+
−1 .
r
r2
r2
r
r2
3
If we multiply by r4 we get
3
2 2 2
2
t2 + r2
t −1
t
+
= 2
⇒ (t2 + r2 )2 = 4t2 r2 + (t2 − r2 )2
2
2
r
r
r
2
2 2
2 2
2
(t + r ) = 4t r + (t − r2 )2 ⇒ z = t2 + r2 , x = 2tr, y = t2 − r2 .
General proof of FLT for exponents of the
form 4n
Theorem 3.1 (FLT for 4v). If v ∈ N, then the equation x4v + y 4v = z 4v
has no solution in integers Z+ − {0}.
Proof.
1. Suppose that the equation x4v + y 4v = z 4v has a solution in integers
Z − {0}.
2. We have x4v + y 4v = z 4v ≡ (x2v )2 + (y 2v )2 = (z 2v )2 , according to one of
the properties of exponents.
3. x2v = (2u)2v = 22v u2v = 2 ∗ 22v−1 u2v ∗ 1, y 2v = 24v−2 u4v − 1, according
to (1) and statement 2.
4. y 2v = 24v−2 u4v − 1 ⇒ 24v−2 u4v = y 2v + 1 ⇒ y = 2m + 1
5. 24v−2 u4v = (2m + 1)2v + 1 = (2m)2v + 2v(2m)2v−1 + · · · + 2v(2m) + 2,
according to statement 4.
24v−2 u4v = 4m(22v−2 m2v−1 + v ∗ 22v−2 m2v−2 + · · · + v) + 2
6. If we divide by 2 we get
24v−3 u4v = 2m(22v−2 m2v−1 +v ∗22v−2 m2v−2 +· · ·+v)+1 ⇒ even = odd.
7. Statement 1 is absurd, according to statement 6.
4
4
General proof of FLT by applying the
general proof for exponents that are
multiples of 4
Proof.
1. (x4v + y 4v = z 4v ) ≡ (x4 )v + (y 4 )v = (z 4 )v
2. Let X = x4 , Y = y 4 , Z = z 4 .
3. (x4v + y 4v = z 4v ) ≡ (X v + Y v = Z v ), according to statements 1 and 2.
4. Let v = 2m or v = 2m + 1, FLT for 4v is proved.
5. (X v + Y v = Z v ) ≡ (X 2m + Y 2m = Z 2m ), where v = 2m, has no solution
in integers Z+ − {0}, according to statement 4.
6. (X v + Y v = Z v ) ≡ (X 2m+1 + Y 2m+1 = Z 2m+1 ) with v = 2m + 1 has no
solution in integers Z+ − {0}, according to statement 4.
Theorem 4.1 (FLT for odd exponents). If n is odd, then the equation
x + y n = z n has no solution in integers Z+ − {0}.
n
Note: The expressions M 4, M 0 4, and M 00 4 represent multiples of 4.
Proof.
1. Suppose that for odd n the equation xn + y n = z n has a solution in
integers Z − {0}.
2. The number n is odd and xn + y n = z n ≡ (xn/2 )2 + (y n/2 )2 = (z n/2 )2 ,
according to one of the properties of exponents.
3. xn/2 = 2n un v n = 2 ∗ 2n−1 un ∗ v n , y n/2 = 22(n−1) u2n − v 2n , 2u and v
coprime numbers, according to (1) and statement 2.
4. y n = (22(2n−1) u2n − v 2n )2 ⇒ v = 2h + 1, according to statement 3.
5. v 2n = (2h + 1)2v = (2h)2v + 2v(2h)2v−1 + · · · + 2v(2h) + 1 = M 0 4 + 1,
according to statement 4.
6. y n = (M 4 − M 0 4 − 1)2 = (M 00 4 − 1)2 , according to statement 5.
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7. y n = (M 00 4 − 1)2 ⇒ y n = (4m − 1)n , according to statement 6.
8. M 4 + 1 = M 00 4 − 1 (n is odd), according to statement 7.
9. According to statement 8 we have M 4 = M 00 4 − 2, which is absurd.
10. Statement 1 is false according to statement 9.
References
W. J. LeVeque. Elementary Theory of Numbers. Addison-Wesley
Publishing Company, 1962.
www.matematicainsolita.8m.com. Pit´agoras, Camino de la Ra´ız.
www.matematicainsolita.8m.com. Pit´agoras, Camino del Binomio.
´
www.matematicainsolita.8m.com. Ultimo
Teorema de Fermat.
viXra:1311.0153. Elementary Proof of Fermat’s Last Theorem and Beal’s
Conjecture.
Author: Carlos Giraldo Ospina, Lic. Matem´aticas, USC, Cali, Colombia
e-mail: [email protected]
Translation and typesetting by: Germ´an Paz
e-mail: germanpaz [email protected]
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