Difference between revisions of "1979 USAMO Problems/Problem 1"
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Determine all non-negative integral solutions <math>(n_1,n_2,\dots , n_{14})</math> if any, apart from permutations, of the Diophantine Equation <math>n_1^4+n_2^4+\cdots +n_{14}^4=1599</math>. | Determine all non-negative integral solutions <math>(n_1,n_2,\dots , n_{14})</math> if any, apart from permutations, of the Diophantine Equation <math>n_1^4+n_2^4+\cdots +n_{14}^4=1599</math>. | ||
| − | == Solution | + | == Solution == |
| + | Recall that <math>n_i^4\equiv 0,1\bmod{16}</math> for all integers <math>n_i</math>. Thus the sum we have is anything from 0 to 14 modulo 16. But <math>1599\equiv 15\bmod{16}</math>, and thus there are no integral solutions to the given Diophantine equation. | ||
| + | |||
{{alternate solutions}} | {{alternate solutions}} | ||
| − | + | == See Also == | |
| − | |||
| − | == See | ||
{{USAMO box|year=1979|before=First Question|num-a=2}} | {{USAMO box|year=1979|before=First Question|num-a=2}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] | ||
Revision as of 10:56, 5 October 2012
Problem
Determine all non-negative integral solutions 
if any, apart from permutations, of the Diophantine Equation 
.
Solution
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1979 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
