Difference between revisions of "1979 USAMO Problems/Problem 1"

(Solution 2)
(Solution 2)
Line 9: Line 9:
 
== 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})=<(\dfrac{1599}{14})^{7/2}<2</math>. So <math>n_1=...=n_{14}=1</math>, clearly a contradiction. Thus there are no solutions to the given equation.
+
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(\dfrac{1599}{14})^{7/2}<2</math>. So <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 14:26, 12 August 2012

Problem

Determine all non-negative integral solutions $(n_1,n_2,\dots , n_{14})$ if any, apart from permutations, of the Diophantine Equation $n_1^4+n_2^4+\cdots +n_{14}^4=1599$.

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 $n_i^4\equiv 0,1\bmod{16}$ for all integers $n_i$. Thus the sum we have is anything from 0 to 14 modulo 16. But $1599\equiv 15\bmod{16}$, and thus there are no integral solutions to the given Diophantine equation.

Solution 2

By AM-GM, $\dfrac{n^{4}_1+...+n^{4}_{14}}{14}\geq(n_1...n_{14})^{2/7}$. Assume there exist $n_i$ which satisfy the given equation. Then $(n_1...n_{14})\leq(\dfrac{1599}{14})^{7/2}<2$. So $n_1=...=n_{14}=1$, clearly a contradiction. Thus there are no solutions to the given equation.

See also

1979 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions