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

Latest revision as of 03:45, 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

@@ Line 2: / Line 2: @@
 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 ==
+== Solution  1==
-{{alternate solutions}}
+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.
+== 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.
-== Solution 2 ==
+== Video Solution by OmegaLearn==
+https://youtu.be/zfChnbMGLVQ?t=4778
-By AM-GM, <math>(n_1...n_{14})^{2/7}=<\dfrac{n^{4}_1+...+n^{4}_{14}}{14}</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.
+~ pi_is_3.14
-== See also ==
+== See Also ==
 {{USAMO box|year=1979|before=First Question|num-a=2}}
+{{MAA Notice}}
 [[Category:Olympiad Number Theory Problems]]

1979 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
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All USAMO Problems and Solutions

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