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 1== |
− | {{ | + | 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. |
− | + | == Solution 2== | |
+ | In base <math>16</math>, this equation would look like: | ||
+ | <cmath>n_1^4+n_2^4+\cdots +n_{14}^4=63F_{16}</cmath> | ||
+ | |||
+ | We notice that the unit digits of the LHS of this equation should equal to <math>F_{16}</math>. In base <math>16</math>, the only unit digits of fourth powers are <math>0</math> and <math>1</math>. Thus, the maximum of these <math>14</math> terms is 14 <math>1's</math> or <math>E_{16}</math>. Since <math>E_{16}</math> is less than <math>F_{16}</math>, there are no integral solutions for this equation. | ||
+ | |||
+ | == Video Solution by OmegaLearn== | ||
+ | https://youtu.be/zfChnbMGLVQ?t=4778 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
− | == See | + | == See Also == |
{{USAMO box|year=1979|before=First Question|num-a=2}} | {{USAMO box|year=1979|before=First Question|num-a=2}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 04:45, 21 January 2023
Problem
Determine all non-negative integral solutions if any, apart from permutations, of the Diophantine Equation
.
Solution 1
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
In base , this equation would look like:
We notice that the unit digits of the LHS of this equation should equal to . In base
, the only unit digits of fourth powers are
and
. Thus, the maximum of these
terms is 14
or
. Since
is less than
, there are no integral solutions for this equation.
Video Solution by OmegaLearn
https://youtu.be/zfChnbMGLVQ?t=4778
~ pi_is_3.14
See Also
1979 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.