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}\geq(n_1...n_{14})^{2/7}</math>. Assume there exist <math>n_i</math> which satisfy the given equation. Then <math>(n_1...n_{14})\leq | + | By AM-GM, <math>\dfrac{n^{4}_1+...+n^{4}_{14}}{14}\geq(n_1...n_{14})^{2/7}</math>. Assume there exist positive integers <math>n_i</math> which satisfy the given equation. Then <math>(n_1...n_{14})\leq{\dfrac{1599}{14}}^{7/2}<2</math>. So <math>n_1...n_{14}=1</math> and <math>n_1=...=n_{14}=1</math>, clearly a contradiction. Thus there are no solutions to the given equation. |
== See also == | == See also == | ||
Revision as of 13:30, 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 positive integers which satisfy the given equation. Then 
. So 
and 
, clearly a contradiction. Thus there are no solutions to the given equation.
See also
| 1979 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
