The Erdos-Strauss conjecture The proof
Jamel Ghannouchi
To cite this version:
Jamel Ghannouchi. The Erdos-Strauss conjecture The proof. Bulletin of Mathematical Sciences
and Applications (www.bmsa.us), 2014, pp.4. <hal-01075686v4>
HAL Id: hal-01075686
https://hal.archives-ouvertes.fr/hal-01075686v4
Submitted on 19 Jan 2015
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The Erdos conjecture
Jamel Ghanouchi
[email protected]
Abstract
(MSC=11D04) In this document, we deal with the Erdos conjecture.
Keywords : Erdos ; Conjecture.
The proof
The Erdos conjecture states that
∀n > 3 ∈ N, ∃x, y, z ∈ N|
4
1 1 1
= + +
n
x y z
There will be no many calculus here, only formulas.
1
1
1
4
=
+
+
6k
3k 3k + 1 3k(3k + 1)
4
1
1
1
=
+
+
6k + 2
6k + 2 2k + 1 (2k + 1)(6k + 2)
1
1
1
4
=
+
+
6k + 3
6k + 3 2k + 2 (2k + 2)(2k + 1)
4
1
1
1
=
+
+
6k + 4
3k + 2 3k + 3 (3k + 3)(3k + 2)
1
1
1
4
=
+
+
6k + 5
6k + 5 2(k + 1) 2(k + 1)(6k + 5)
And
1
+
k+c
=
1
b(6k+1)(k+c)
(k+c)(4b−1)−(6k+1)b
1
(1 +
k+c
1
b(6k+1)
(k+c)(4b−1)−b(6k+1)
1
)
=
(k + c)(4b − 1) − b(6k + 1)
1
(1 +
)
k+c
b(6k + 1)
=
=
Thus
1 (k + c)(4b − 1)
(
)
k+c
b(6k + 1)
4b − 1
4
1
=
−
b(6k + 1)
6k + 1 b(6k + 1)
1
1
4
=
+
+
6k + 1
b(6k + 1) k + c
1
b(6k+1)(k+c)
(k+c)(4b−1)−(6k+1)b
b(6k+1)(k+c)
We choose b, c such as (k+c)(4b−1)−(6k+1)b
is a positive integer and it is always
possible ! For example
k=1⇒
7b(1 + c)
b(6k + 1)(k + c)
=
(k + c)(4b − 1) − (6k + 1)b
(1 + c)(4b − 1) − 7b
7b(1 + c)
4bc − 3b − c − 1
c = 1, b = 3 ⇒ 4bc − 3b − c − 1 = 1
1 1
1
4
⇒ = + +
7
7 2 14
b(6k + 1)(k + c)
13b(2 + c)
k=2⇒
=
(k + c)(4b − 1) − (6k + 1)b
(2 + c)(4b − 1) − 13b
=
13b(2 + c)
4bc − 5b − c − 2
c = 2, b = 2 ⇒ 4bc − 5b − c − 2 = 2
4
1
1
1
⇒
=
+ +
13
26 4 52
b(6k + 1)(k + c)
19b(3 + c)
k=3⇒
=
(k + c)(4b − 1) − (6k + 1)b
(3 + c)(4b − 1) − 19b
=
19b(3 + c)
4bc − 7b − c − 3
c = 3, b = 6 ⇒ 4bc − 7b − c − 3 = 1
4
1
1
1
⇒
=
+ +
19
114 6 684
Etc... Untill infinity, it means that it is always possible to have three egyptian fractions !
=
Conclusion
By series of formulas, we have proved the Schinzel conjecture.
2
Références
[1] Yamamoto Koichi, On the diophantine equation 4/n = 1/c+1/y +1/z
Memoirs of the Faculty of Science , (1965).
3