Difference between revisions of "1979 USAMO Problems/Problem 1"
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== Solution 2 == | == Solution 2 == | ||
− | By AM-GM, <math>\dfrac{n^{4}_1+...+n^{4}_{14}}{14}>=(n_1...n_{14})^{2/7}</math>. Assume there exist <math>n_i</math> which satisfy the given equation. Then | + | By AM-GM, <math>\dfrac{n^{4}_1+...+n^{4}_{14}}{14}>=(n_1...n_{14})^{2/7}</math>. Assume there exist <math>n_i</math> which satisfy the given equation. Then <math>(\dfrac{1599}{14})^{7/2}>=(n_1..n_{14})</math> |
== See also == | == See also == |
Revision as of 14:16, 12 August 2012
Contents
Problem
Determine all non-negative integral solutions if any, apart from permutations, of the Diophantine Equation
.
Solution 1
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Recall that for all integers
. Thus the sum we have is anything from 0 to 14 modulo 16. But
, and thus there are no integral solutions to the given Diophantine equation.
Solution 2
By AM-GM, . Assume there exist
which satisfy the given equation. Then
See also
1979 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |