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 9: / Line 9: @@
 <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 Also ==

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

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