(Erdös - Strauss Conjecture : D11 Egyptian fractions)

- Erdös and Strauss conjectured that the Diophantine equation
- In this chapter, I will show,
- when b=kn, we can construct a parametrized solution by the solution n, a, b, c.
- when b≠kn, we can construct a parametrized solution by the solution n, a, b, c in heuristic way.
- In order to understand the difficulty of this problem, please try to search the solutions of following Diophantine equations.
- Q1 : 4/11=1/a+1/b+1/c
- Q2 : 5/11=1/a+1/b+1/c
- Q3 : 6/11=1/a+1/b+1/c
- Q4 : 7/11=1/a+1/b+1/c
- Q5 : 8/11=1/a+1/b+1/c

could be solved in positive integers for all n < 1. (still not be solved.)4/n=1/a+1/b+1/c

It is checked n≤10

for the denominator is 4, 5, 6, 7 and any integer.

Applying this method, we can extend the search range dramatically.

- How to solve this equation
- Program
- Table of solutions (computation results)
- Parametrized solutions which contain the solution a=b or b=c
- How to construct the parametrized solution whic contains the solution b=kp
- Tables of the parametrized solutions of the case b=kp
- The case that the solution could not find under a≤100
- How to construct the parametrized solution from each solution (complete)
- How to construct the parametrized solution from the solution b≠kp in heuristic way
- Program
- Table of solutions and parametrized solutions
- 4/p=1/a+1/b+1/c, final (2`47) , (53`97)
- 5/p=1/a+1/b+1/c, final (2`47) , (53`97)
- 6/p=1/a+1/b+1/c, final (2`47) , (53`97)
- 7/p=1/a+1/b+1/c, final (3`47) , (53`97)
- Conclusion

Chapter 0 Introduction |
"Mathematician's Secret Room" | Chapter 2 Squares consisted of 3 different digits |
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Chapter 0 (Japanese) | index (Japanese) | Chapter 2 (Japanese) |

E-mail : kc2h-msm@asahi-net.or.jpHisanori Mishima