Feb 10, 2018 - Abstract. The Fermat's Last Theorem states that the equation: xn + yn = zn ... which is equal to Lucas n-th Cyclic Polynomials. Then this sum ...
On Fermat’s Last Theorem Gevorg Hmayakyan February 10, 2018 Abstract The Fermat’s Last Theorem states that the equation: xn + y n = z n has no integer solutions for n ≥ 3. It can be restated for rational A and B as: An + B n = 1 has no rational solutions for n ≥ 3. In this paper the non existence of solution is proved for prime powers p > 5. The prove is based on rational number pair representation as √ k ± k2 + 4m 2 More accurately for any rational numbers A√and B there is an according √ k− k2 +4m k+ k2 +4m rational pair k and m for which the A = and B = . 2 √2 k− k2 +4m The first part is investigation of n-th powers sum of the 2 √ k+ k2 +4m which is equal to Lucas n-th Cyclic Polynomials. and 2 Then this sum equality to 1 is investigated. √ And finally using the rationality condition for k2 + 4m the main result is proven. Which states that solution of Fermat’s Last Theorem is equivalent to un + z n = (u + 1)n solution existence in integers. And as a final step the Fermat’s Last Theorem is proven for primes p > 5 by proving the unsolvability of last equation.
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1
Basic Lemmas
Lemma 1. Any pair of rational √numbers A and B can be represented in the √ 2 2 form A = k− k2 +4m and B = k+ k2 +4m , where the k and m are also rational √ numbers with obviously rational k 2 + 4m Proof. Let take k = A + B and m = −AB. This solves the equation system obviously and as the A and B are rational then m and k are rational appropriately. Lemma 2. k−
√
k 2 + 4m 2
!n +
k+
√
k 2 + 4m 2
!n = Ln (k, m)
where the Ln (k, m) is Lucas Cyclic Polynomial. Lemma 3. For prime p the p−1
Lp (k, m) = k
2 X
αpi k p−1−2i mi
i=0 p−1
where αp0 = 1 and for other coefficients p|αpk , especially the αp1 = αp 2 = p Lemma 4. For prime p the Lp
p−1 2 X p 1 √ ,1 x2 = αpi xi x i=0
Lemma 5. For prime p and x = u(u + 1) the p 1 Lp √ , 1 x 2 = (u + 1)p − up x Lemma 6. The equation 1 + 4x = y 2 has solution if x = u(u + 1)
2
2
Main Result
Theorem 1. If m, k and
√
k 2 + 4m are rational then for prime p > 5 from the Lp (k, m) = 1
follow k=
1 m1 , m = 2 , m1 = u(u + 1) > 0 k2 k2
And
p 1 p √ , 1 m12 = k2 m1
Lp Proof. Assume
Lp (k, m) = 1 m1 m2 ,
The k = kk12 gcd(m1 , m2 )
and m = where k1 , k2 , m1 , m2 are integers and gcd(k1 , k2 ) = = 1. And also for negative m it is assumed that m1 is negative and the m2 is positive. According to Lemma 3 p−1
p−1−2i i 2 k1 X k1 m1 i α =1 k2 i=0 p k2 m2 p−1
By multiplying both sides on k2p m2 2 : p−1
k1
2 X
p−1
αpi k1p−1−2i k22i mi1 m2 2
−i
p−1
= k2p m2 2
i=0
From Lemma 3 p−3 p−1
k2p m2 2
p−1
=
k1 (k1p−1 m2 2
p−3
+pk1p−3 k22 m1 m2 2
+
2 X
p−1
αpi k1p−1−2i k22i mi1 m2 2
−i
p−1
+pk2p−1 m1 2 )
i=2
as the gcd(k1 , k2 ) = 1 m2 = k1 m21 and: k2p (k1 m21 )
p−1 2
= k1 (k1p−1 (k1 m21 )
p−1 2
+pk1p−3 k22 m1 (k1 m21 )
p−3 2
P p−3 p−1 2 + i=2 αpi k1p−1−2i k22i mi1 (k1 m21 ) 2 −i +
p−1
pk2p−1 m1 2 ) By dividing both sides on k1 and simplifying: p−3 p−3 2
k2p k1
p−1 2
m21
3(p−1) 2
= k1
p−1 2
3(p−3) 2
m21 +pk1
p−3 2
k22 m1 m21 +
2 X
i=2
3
3(p−2i−1) 2
αpi k1
p−1
k22i mi1 m212
−i
p−1
+pk2p−1 m1 2
Using simple transformation: p−3 3(p−1) 2
k1
p−1 2
m21
p−3
=
3(p−3) 2
p−1 2
k22 (k2p−2 k1 2
m21 −(pk1
p−3 2
m1 m21 +
2 X
3(p−2i−1) 2
αpi k1
p−1
k22i−2 mi1 m212
−i
p−1
+pk2p−3 m1 2 ))
i=2
From this and gcd(k1 , k2 ) = 1 follows: k2 |m21 . So m21 = k2 m22 3(p−1) 2
p−3 2
p−1 2
3(p−3)
p−1
k1 (k2 m22 ) = k22 (k2p−2 k1 (k2 m22 ) 2 − (pk1 2 m1 (k2 m22 ) 3(p−2i−1) p−1 P p−3 p−1 i 2 2 k22i−2 mi1 (k2 m22 ) 2 −i + pk2p−3 m1 2 )) i=2 αp k1
p−3 2
+
By simplifying and collecting k2 : p−3 3(p−1) 2
k1
p−1 2
k2
p−1 2
m22
p+1 2
= k2
p−3
(k2p−1 k1 2
3(p−3) 2
p−1 2
m22 −pk1
p−3 2
m1 m22 −
2 X
3(p−2i−1) 2
αpi k1
p−1
k2i mi1 m222
−i
p−3
i=2 p−1
By dividing both sides on k2 2 and simplifying: p−3 3(p−1) 2
k1
p−1 2
m22
p−3
=
k2 (k2p−1 k1 2
3(p−3) 2
p−1 2
m22 −pk1
p−3 2
m1 m22
2 X
3(p−2i−1) 2
αpi k1
p−1
k2i−1 mi1 m222
−i
p−3
p−1
−pk2 2 m1 2 )
i=2
And again as the gcd(k1 , k2 ) = 1 m22 = k2 m23 3(p−1) 2
p−3 2
p−1
3(p−3)
p−1
k1 (k2 m23 ) 2 = k2 (k2p−1 k1 (k2 m23 ) 2 − pk1 2 m1 (k2 m23 ) 3(p−2i−1) p−3 p−1 P p−3 p−1 i 2 2 k2i−1 mi1 (k2 m23 ) 2 −i − pk2 2 m1 2 ) i=2 αp k1
p−3 2
−
After simplifying: 3(p−1) 2
k1
p−3
k2 2
p−1
3(p−1)
p−1
p−3
3(p−3)
p−1
p−3
p−3
k2 2 m232 = k2 (k2 2 k1 2 m232 − pk1 2 k2 2 m1 m232 − 3(p−2i−1) p−1 p−3 p−1 P p−3 −i i 2 2 mi1 m232 − pk2 2 m1 2 ) i=2 αp k1
and collecting k2 : p−3 3(p−1) 2
k1
p−1 2
k2
p−1 2
m23
p−1 2
= k2
p−3 2
(k1
3(p−3) 2
p−1 2
k2p m23 −pk1
p−3 2
m1 m23 −
2 X
3(p−2i−1) 2
αpi k1
p−1
mi1 m232
−i
i=2 p−1 2
dividing on k2
p−3 3(p−1) 2
k1
p−1 2
m23
p−3 2
= k2p k1
p−1 2
3(p−3) 2
m23 −pk1
p−3 2
m1 m23 −
2 X
i=2
4
3(p−2i−1) 2
αpi k1
p−1
mi1 m232
−i
p−1
−pm1 2
p−1
−pk2 2 m1 2 )
p−1
−pm1 2 )
By collecting k1 : p−5
p−1
3(p−3)−2 2
p−1
pm1 2 = k1 (k2p k1 2 m232 −pk1 3(p−1)−2 2
k1
p−3
m1 m232 −
P p−3 2 i=2
3(p−2i−1)−2 2
αpi k1
p−1
mi1 m232
−i
−
p−1 2
m23 )
Obviously k1 can not divide m1 as it divides m2 and gcd(m1 , m2 ) = 1. So either k1 = 1 or k1 = p. Case k1 = 1: p−3 p−1 2
pm1
p−1
=
2 X
p−3 2
k2p m232
− pm1 m23 −
p−1
αpi mi1 m232
−i
p−1
− m232
i=2
By collecting m23 : p−3 p−1 2
pm1
p−3
=
m23 (k2p m232
p−5 2
− pm1 m23 −
2 X
p−1
αpi mi1 m232
−i−1
p−3
− m232 )
i=2
from this and gcd(m23 , m1 ) = 1 follows that either the m23 = 1 or m23 = p. Sub-case k1 = 1 and m23 = 1: p−3 p−1 2
pm1
=
k2p
− pm1 −
2 X
αpi mi1 − 1
(1)
i=2
Sub-case k1 = 1 and m23 = p: p−1 2
pm1
= p k2p p
p−3 2
p−3
− pm1 p
p−5 2
−
2 X
αpi mi1 p
p−1 2 −i−1
−p
p−3 2
i=2
or
p−3 p−1 2
m1
=
p−3 k2p p 2
− pm1 p
p−5 2
−
2 X
αpi mi1 p
p−1 2 −i−1
−p
p−3 2
i=2
So the p divides m1 as the Lemma 3 states that p|αpi , which is contradiction as the p divides m2 and gcd(m1 , m2 ) = 1. Case k1 = p: p−1
pm1 2 = p(k2p p p
3(p−1)−2 2
p−5 2
p−1
m232 −pp
3(p−3)−2 2
p−3
m1 m232 −
p−1 2
m23 )
or 5
P p−3 2 i=2
αpi p
3(p−2i−1)−2 2
p−1
mi1 m232
−i
−
p−1
m1 2 = k2p p p
3(p−1)−2 2
p−5 2
p−1
m232 − pp
3(p−3)−2 2
p−3
m1 m232 −
P p−3 2 i=2
αpi p
3(p−2i−1)−2 2
p−1
mi1 m232
−i
−
p−1 2
m23
or by collecting m23 : p−1
m1 2 = m23 (k2p p p
3(p−1)−2 2
p−5 2
p−3
m232 −p
3(p−3) 2
p−5
m1 m232 −
P p−3 2 i=2
αpi p
3(p−2i−1)−2 2
p−1
mi1 m232
−i−1
−
p−3 2
m23 )
from here follows m23 = 1 as the gcd(m1 , m23 ) = 1: p−3 p−1 2
m1
=
p−5 k2p p 2
−p
3(p−3) 2
m1 −
2 X
αpi p
3(p−2i−1)−2 2
mi1 − p
3(p−1)−2 2
i=2
As the p > 5 from above follows that p divides m1 which is contradiction as the k1 = p divides m2 . Now lets consider (1). For this case: m2 = k1 k22 m23 and according to the case: k1 = 1, m23 = 1 so k1 = 1, m2 = k22 Now by taking into account the rationality condition k = k1 /k2 and m = m1 /m2 : k 2 + 4m =
1 k2
2
or 1 + 4m1 =
+4
m1 r12 = k22 r22
k22 r12 r22
from this obviously 1 + 4m1 = N 2 where N is integer. According to Lemma 6 m1 = u(u + 1) So k=
1 u(u + 1) ,m = k2 k22 6
√
k 2 + 4m and also
Now by reordering the (1): p−3 p−1 2
pm1
+ pm1 +
2 X
αpi mi1 + 1 = k2p
i=2
and taking into account Lemma 3: p−3 p−1 2
pm1
+ pm1 +
2 X
p−1
αpi mi1
+1=
i=2
2 X
αpi mi1
i=0
According to Lemma 4: p−1 2 X
αpi mi1 = Lp
i=0
p 1 √ , 1 m12 m1
So Lp
p 1 p √ , 1 m12 = k2 m1
The Theorem 1 is proved. From the Theorem 1 immediately follows: Lemma 7. If Lp (k, m) = 1 and k > 0 then m > 0 Theorem 2. The equation: Ap + B p = 1 for rational A, B and prime p > 5 has no solution if A and B are positive, otherwise it has solution only if has solution the xp + y p = (x + 1)p for integer x and y Proof. According to Lemma 1 A and B can be represented as √ √ k + k 2 + 4m k − k 2 + 4m ,B = A= 2 2 √ with condition of rationality of k 2 + 4m According to Lemma 2 the Ap + B p = Lp (k, m) = Lp (A + B, −AB). From Lemma 7 follows that if both A and B are positive, then Lp (A + B, −AB) = 1 7
equation has no solution. Considering the case where −AB and A + B both are positive. According to Theorem 1 the Ap + B p = 1 may have solution if: k = k12 , m = u(u+1) and k22 p 1 , 1 m12 = k2p , where m1 = u(u + 1). Lp √m 1 According to Lemma 5: Lp
p 1 √ , 1 m12 = Lp m1
!
1 p
p
u(u + 1)
, 1 (u(u + 1)) 2 = (u + 1)p − up
so (u + 1)p − up = k2p The Theorem 2 is proved. Theorem 3. The equation: xp + y p = (x + 1)p has no solution for integer x and y Proof. Obviously |y| < |x|. By dividing both sides on (x + 1)p : p p x y + =1 x+1 x+1 Now solution of: x k− = x+1
√
is: k=
k 2 + 4m y k+ , = 2 x+1
√
k 2 + 4m 2
xy x+y ,m = − x+1 (x + 1)2
Assume x + y = rt1 and x + 1 = rt2 with gcd(t1 , t2 ) = 1. Then the Theorem 1 states that t1 + t2 = 1. Or 2x + y + 1 = r So: x + 1 = (2x + y + 1)t2 This can not take place as the |y| < |x| and the |2x + y + 1| > |x + 1| accordingly.
3
Conclusion
The Theorem 2 and Theorem 3 prove the Fermat Last Theorem for prime p > 5.
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References [1] Tewodros Amdeberhan, Mahir Bilen Can, and Melanie Jensen: Divisors and specializations of Lucas polynomials, https://arxiv.org/pdf/1406.0432.pdf [2] Koshy, T: Fibonacci and Lucas Numbers with Applications. New York: Wiley, 2001 [3] Sloane, N. J. A.: Sequence A002378 in ”The On-Line Encyclopedia of Integer Sequences.” https://oeis.org/A002378 [4] Sloane, N. J. A.: Sequence A114525 in ”The On-Line Encyclopedia of Integer Sequences.” https://oeis.org/A114525
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