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| Title: | Mathematics at DEC |
|
| Moderator: | RUSURE::EDP |
|
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
1787.0. "College Mathematics Journal #509" by RUSURE::EDP (Always mount a scratch monkey.) Fri Sep 10 1993 14:26
Proposed by Norman Schaumberger, Hofstra University, Hempstead, NY.
Prove or disprove that, for each positive integer n, the diophantine
equation
x^n + y^(n+1) + z^(n+2) + w^(n+3) = v^(n^2+3n+1)
has a solution in positive integers.
| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1787.1 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Sat Sep 11 1993 22:30 | 19 |
| Let
P = n(n+1)(n+2)(n+3)
x = 4^(P/n)
y = 4^(P/(n+1))
z = 4^(P/(n+2))
w = 4^(P/(n+3))
Then the left hand side is 4(4^P) = 4^(P+1) which is
4^(n^4 + 6n^3 + 11n^2 + 6n + 1) = 4^((n^2 + 3n + 1)(n^2 + 3n + 1))
= (4^(n^2 + 3n + 1))^(n^2 + 3n + 1). So you can combine the
above with
v = 4^(n^2 + 3n + 1)
to get a solution in positive integers for any positive
integer n.
Dan
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