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

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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 0-14 mod16. But 1599=15mod16, and thus there are no integral solutions to the given Diophantine equation.
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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.
  
 
== See also ==
 
== See also ==

Revision as of 08:45, 13 May 2011

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

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.

See also

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