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 1 ==
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== Solution 1==
{{alternate solutions}}
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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.
  
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.
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== Solution  2==
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In base <math>16</math>, this equation would look like:
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<cmath>n_1^4+n_2^4+\cdots +n_{14}^4=63F_{16}</cmath>
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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.
  
== Solution 2 ==
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== Video Solution by OmegaLearn==
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https://youtu.be/zfChnbMGLVQ?t=4778
  
By AM-GM, <math>\dfrac{n^{4}_1+...+n^{4}_{14}}{14}\geq(n_1...n_{14})^{2/7}</math>. Assume there exist nonnegative 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}=0</math> or <math>n_1...n_{14}=1</math>, clearly a contradiction. Thus there are no solutions to the given equation.
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~ pi_is_3.14
  
== See also ==
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== See Also ==
 
{{USAMO box|year=1979|before=First Question|num-a=2}}
 
{{USAMO box|year=1979|before=First Question|num-a=2}}
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{{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 $(n_1,n_2,\dots , n_{14})$ if any, apart from permutations, of the Diophantine Equation $n_1^4+n_2^4+\cdots +n_{14}^4=1599$.

Solution 1

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

In base $16$, this equation would look like: \[n_1^4+n_2^4+\cdots +n_{14}^4=63F_{16}\]

We notice that the unit digits of the LHS of this equation should equal to $F_{16}$. In base $16$, the only unit digits of fourth powers are $0$ and $1$. Thus, the maximum of these $14$ terms is 14 $1's$ or $E_{16}$. Since $E_{16}$ is less than $F_{16}$, 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 (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

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